## Presentation on theme: "Solving Quadratic Equations Solving Quadratic Equations"— Presentation transcript:

Quadratic Equations and Problem Solving This module explores different methods for solving quadratic equations. The stated goal is the finding of the solution set for the equation. We emphasize that while the equation might or might not have a solution in the real numbers, it always has a solution set, possibly the empty set. Problems are set up to produce various types of equations as examples. Quadratic Equations 7/9/2013

Solutions A solution for the equation ax2 + bx + c = 0 is a value of variable x satisfying the equation NOTE: For now we assume all values are real numbers The solution set for the equation is the set of all of its solutions Quadratic Functions: Solutions As with linear equations we define solutions for quadratic equations as the values of x that satisfy the equation. The examples simply show the solutions and solution sets. Note that one quadratic has two solutions, another has one solution, and the last equation has no solution. Note that all of the equations have solution sets even if there are no solutions. The empty set provides an elegant and very simple way of saying there are no solutions. Remember: an equation won’t always have a solution but it will always have a solution set. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Examples 1. x2 + 2x – 3 = 0 2. 16x2 – 8x + 1 = 0 3. x2 + 2x + 5 = 0 Solutions: –3, 1 Solution set: {–3 , 1} Solution set: { } 1 4 1 4 Solution: Quadratic Functions: Solutions As with linear equations we define solutions for quadratic equations as the values of x that satisfy the equation. The examples simply show the solutions and solution sets. Note that one quadratic has two solutions, another has one solution, and the last equation has no solution. Note that all of the equations have solution sets even if there are no solutions. The empty set provides an elegant and very simple way of saying there are no solutions. Remember: an equation won’t always have a solution but it will always have a solution set. Solutions: none Solution set: { } OR  7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Square Root Property
Solving Quadratic Equations Reduced Standard Form - Type 1 ax2 = c , for a ≠ 0 Rewrite: Square Root Property If x2 = k then c a x2 = = k The Square Root Property We recognize the fact that perfect squares can have two square roots and use this fact to solve simple quadratic equations. A very common mistake made by students is to assume that somehow has two values, when in fact it is simply a number, and in fact a non-negative number, if we are to stay in the real number system. If we wish to identify a negative number whose square is k we refer to , whose square is most certainly equal to k. Confusion on this points leads students to sometimes say such things as or , neither of which is true. We point out here that a basic concept involving roots is the concept of a principal root. When finding the two possible roots of a square the positive root is always the principal root, designated here as The second square root is always the negative of the principal root, designated here as In the illustration the statement that x2 = k tells us that either x = or that x = , but not both. If we think about finding the square root of x2, that is , then we can think of this as From the definition of absolute value we then have or Recall that for any real number a, a , for a ≥ 0 – a , for a < 0 k = x x2 = x │ │ OR k = x 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Square Root Property
Solving Quadratic Equations Square Root Property If x2 = k then k = x x2 = x │ │ OR k = x Note: This does NOT say that  2 4 = The Square Root Property We recognize the fact that perfect squares can have two square roots and use this fact to solve simple quadratic equations. A very common mistake made by students is to assume that somehow has two values, when in fact it is simply a number, and in fact a non-negative number, if we are to stay in the real number system. If we wish to identify a negative number whose square is k we refer to , whose square is most certainly equal to k. Confusion on this points leads students to sometimes say such things as or , neither of which is true. We point out here that a basic concept involving roots is the concept of a principal root. When finding the two possible roots of a square the positive root is always the principal root, designated here as The second square root is always the negative of the principal root, designated here as In the illustration the statement that x2 = k tells us that either x = or that x = , but not both. If we think about finding the square root of x2, that is , then we can think of this as From the definition of absolute value we then have or Recall that for any real number a, a , for a ≥ 0 – a , for a < 0 This DOES say that IF x2 = 4 then = 2 x = 4 OR 4 x = –2 The Principal Root The Negative Root 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Square Root Property Solving Quadratic Equations Roof repairs on a 100-story building A tar bucket slips and falls to the sidewalk If each story is feet, how long does it take the bucket to hit the sidewalk? Distance of fall: (12.96 ft/story)(100 stories) = 1296 ft For a fall of s ft in t seconds we have The Square Root Property: Example In this example we solve a quadratic equation, which eventually leads us to finding a square root of a real (positive) number. We recognize that we can choose either the principal root (the positive value) or the second root (which is negative). Of course we choose the positive root (i.e. the principal root) because it is the one that makes sense in the real-world example we are describing. The equation given is a shortened version of a standard physics equation, in fact one from classical mechanics. For distance s(t) from a reference point traveled in a straight line by an object for time t, the equation is s = ½ at2 + v0t + s0 where a is the acceleration applied during time t, is the velocity at time t, v0 is the initial velocity, and s0 is the initial distance from the reference point. Here s0 = 0, v0 = 0 and a is the acceleration due to gravity, i.e. 32 ft/sec2. Thus, s = ½ (32)t2 + (0)t + 0 = 16t2 . s = 16t2 OR t2 = s/16 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Square Root Property Solving Quadratic Equations Roof repairs on a 100-story building Solving for t We choose the positive value of t Thus s = 16t2 OR t2 = s/16 = t 4 s = t s 4 OR WHY ? The Square Root Property: Example In this example we solve a quadratic equation, which eventually leads us to finding a square root of a real (positive) number. We recognize that we can choose either the principal root (the positive value) or the second root (which is negative). Of course we choose the positive root (i.e. the principal root) because it is the one that makes sense in the real-world example we are describing. The equation given is a shortened version of a standard physics equation, in fact one from classical mechanics. For distance s(t) from a reference point traveled in a straight line by an object for time t, the equation is s = ½ at2 + v0t + s0 where a is the acceleration applied during time t, is the velocity at time t, v0 is the initial velocity, and s0 is the initial distance from the reference point. Here s0 = 0, v0 = 0 and a is the acceleration due to gravity, i.e. 32 ft/sec2. Thus, s = ½ (32)t2 + (0)t + 0 = 16t2 . = t 4 s = 4 1296 4 36 = 9 = seconds 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Special Forms Solving Quadratic Equations Perfect-Square Trinomials Always the square of a binomial A2 + 2AB + B2 = (A + B)2 A2 – 2AB + B2 = (A – B)2 … OR (B – A)2 Special Forms: The Perfect-Square Trinomial It is useful to be able to recognize certain special forms, since they often arise in the solution of problems. The perfect-square trinomial is one of the most common and most useful of the special-form polynomials. It arises from squaring a binomial of form A + B or A – B or B – A, where A and B are any algebraic expressions. In the examples A is successively x, 2y, 3y and xy, while B takes on values 3, 5, 2x and 3z, respectively. Writing out the squares on the right hand side of the equations easily verifies the validity of the identity equations. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Special Forms Solving Quadratic Equations Examples x2 + 6x + 9 4y2 – 20y + 25 9y2 + 12xy + 4x2 x2y2 – 6xyz + 9z2 = (x + 3)2 = (2y – 5)2 = (3y + 2x)2 Special Forms: The Perfect-Square Trinomial It is useful to be able to recognize certain special forms, since they often arise in the solution of problems. The perfect-square trinomial is one of the most common and most useful of the special-form polynomials. It arises from squaring a binomial of form A + B or A – B or B – A, where A and B are any algebraic expressions. In the examples A is successively x, 2y, 3y and xy, while B takes on values 3, 5, 2x and 3z, respectively. Writing out the squares on the right hand side of the equations easily verifies the validity of the identity equations. = (xy – 3z)2 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Completing the Square Solving Quadratic Equations Convert to perfect-square trinomial Solve using square root property Example 1: 2x2 – 3x + 5 = x2 – 9x + 12 x2 + 6x = 7 To make the left side we need to find a such that (x + a)2 = x2 + 2ax + a2 Subtract x2 and 5, on each side, and add 9x a perfect square Completing the Square One useful method for solving quadratic equations uses the perfect-square trinomial form and the square root property. This method is known as the method of completing the square. In the example we first gather on x-terms together on one side of the equation, consolidating all constant terms on the other side of the equation. We note that the coefficient of x is already 1 (so there is no need to divide by it to force the lead coefficient to 1). Using the same technique we used for finding the vertex form of a quadratic, we write the linear term (x but not x2) with a coefficient made up of two factors: 2 and whatever else is left from the original factor, i.e. factor 2 out of the original coefficient. The “other” factor is then squared and added to both sides of the equation. The expression with the x and x2 terms is now a perfect-square trinomial, which we then write as the square of appropriate binomial. This is shown in the illustration as (x + 3)2 with 16 on the other side of the equation. We now invoke the square root property to find that there are two roots equal to the perfect-square trinomial: +4 and -4. Isolating x by subtracting 3 from the x +3 and from each of +4 and -4, we find that there are two values of x (the roots) that can be assigned to x to make the original equation true. Thus the solutions are -7 and 1, so the solution set is { -7, 1 }. = x2 + 6x + a2 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Completing the Square Solving Quadratic Equations Convert to perfect-square trinomial Example 1: 2x2 – 3x + 5 = x2 – 9x + 12 x2 + 6x = 7 (x + a)2 = x2 + 2ax + a2 = x2 + 6x + a2 So 2ax = 6x giving a = 3 and a2 = 9 Completing the Square One useful method for solving quadratic equations uses the perfect-square trinomial form and the square root property. This method is known as the method of completing the square. In the example we first gather on x-terms together on one side of the equation, consolidating all constant terms on the other side of the equation. We note that the coefficient of x is already 1 (so there is no need to divide by it to force the lead coefficient to 1). Using the same technique we used for finding the vertex form of a quadratic, we write the linear term (x but not x2) with a coefficient made up of two factors: 2 and whatever else is left from the original factor, i.e. factor 2 out of the original coefficient. The “other” factor is then squared and added to both sides of the equation. The expression with the x and x2 terms is now a perfect-square trinomial, which we then write as the square of appropriate binomial. This is shown in the illustration as (x + 3)2 with 16 on the other side of the equation. We now invoke the square root property to find that there are two roots equal to the perfect-square trinomial: +4 and -4. Isolating x by subtracting 3 from the x +3 and from each of +4 and -4, we find that there are two values of x (the roots) that can be assigned to x to make the original equation true. Thus the solutions are -7 and 1, so the solution set is { -7, 1 }. x2 + 6x + 9 = 7 + 9 = 16 Adding 9 to each side of x2 + 6x = 7 (x + 3)2 = 16 x + 3 16 = + = + 4 x = –3 + 4 Solution set: { –7, 1 } 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Completing the Square Solving Quadratic Equations Example 2: 5y2 – 87y + 89 = y – 4y2 9y2 – 102y = –64 (3y – 17)2 = 225 3y – 17 225 = + 15 = + 3y = 17 15 + = 32 3 Completing the Square In this example, after gathering and consolidating like powers of y and the constants, we wind up with a negative constant on the right had side of the equation and a lead coefficient of 9. In this case, instead of dividing out the 9 to force the lead coefficient to 1 (which would make some awkward fractions), we incorporate the 9, as the square of 3, into the squared term. This gives us the term (3y)2. So, the new variable is 3y instead of just y. Thus, as before, the linear term must contain a factor of 2 but also a factor of 3y, instead of just y. Factoring out 2 from -102y gives us -51y and factoring out 3y from -51y gives us Squaring -17 yields 289 as the constant term to be added to both sides of the equation. The illustration shows another way to view this process. Knowing the form the perfect-square trinomial must have as the square of (3y + a), we can equate the middle term 6ay to given middle term -102y and solve for a. This is shown in the illustration. The last step is applying the square root property to the resulting equation (3y – 17)2 = 225 and then solving the simple linear equation so obtained. We find two solutions and express them as a solution set. The solutions are expressed as rational numbers, which are both more accurate and more descriptive than decimal fractions. If this problem is part of a larger problem which will use these solutions, greater accuracy can be obtained by leaving the solutions in rational form as long as possible. Final results can always be converted to decimal fractions later. y = + 17 3 15 2 3 Solution Set: { 32 3 2 , } 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Solving Quadratic Equations Complete the Square on the Standard Form ax2 + bx + c = 0 , for a ≠ 0 Rewrite: move c, divide by a c a b x2 x + = Now complete the square on the left side Want a constant k such that The Quadratic Formula We here solve the general quadratic equation in standard form by completing the square. The steps are exactly the same as before: divide out the lead coefficient to force the new lead coefficient to 1 subtract the constant term from both sides of the equation rewrite the linear term with a factor of 2 (multiply and divide by 2) x2 + 2(b/2a)x = -c/a square the resulting cofactor and add it to both sides of the equation x2 + 2(b/2a)x + (b/2a)2 = (b/2a)2 – c/a recognize the perfect-square trinomial thus produced and write it as the square of a binomial x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 = (b/2a)2 – c/a apply the square root property to equate the new binomial to the constant on the right x + b/2a = (   b2 – 4ac )/2a subtract the constant term of the binomial from both sides to isolate x consolidate the constants on the right x = (– b   b2 – 4ac )/2a The result is the well known quadratic formula. Use of this formula will always produce a correct solution (or pair of solutions) – even when the solutions are not real numbers. b a x2 x + + k2 is a perfect square – that is b a x2 x + + k2 = k ( ) 2 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Solving Quadratic Equations Complete the Square on the Standard Form c a b x2 x + = b a x2 x + + k2 = k ( ) 2 = x2 + 2xk + k2 2k = b a = b 2a ( ) 2 k2 Thus … and The Quadratic Formula We here solve the general quadratic equation in standard form by completing the square. The steps are exactly the same as before: divide out the lead coefficient to force the new lead coefficient to 1 subtract the constant term from both sides of the equation rewrite the linear term with a factor of 2 (multiply and divide by 2) x2 + 2(b/2a)x = -c/a square the resulting cofactor and add it to both sides of the equation x2 + 2(b/2a)x + (b/2a)2 = (b/2a)2 – c/a recognize the perfect-square trinomial thus produced and write it as the square of a binomial x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 = (b/2a)2 – c/a apply the square root property to equate the new binomial to the constant on the right x + b/2a = (   b2 – 4ac )/2a subtract the constant term of the binomial from both sides to isolate x consolidate the constants on the right x = (– b   b2 – 4ac )/2a The result is the well known quadratic formula. Use of this formula will always produce a correct solution (or pair of solutions) – even when the solutions are not real numbers. b a x2 x + + b 2a ( ) 2 c a = + b 2a ( ) 2 x + b 2a ( ) 2 c a = 4a 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Solving Quadratic Equations Complete the Square on the Standard Form x + ( b 2a ) 2 = c a + b 2 4a + b 2 4a = 4ac = 4a 2 b 4ac Square Root Property = 4a 2 b 4ac + x + b 2a The Quadratic Formula We here solve the general quadratic equation in standard form by completing the square. The steps are exactly the same as before: divide out the lead coefficient to force the new lead coefficient to 1 subtract the constant term from both sides of the equation rewrite the linear term with a factor of 2 (multiply and divide by 2) x2 + 2(b/2a)x = -c/a square the resulting cofactor and add it to both sides of the equation x2 + 2(b/2a)x + (b/2a)2 = (b/2a)2 – c/a recognize the perfect-square trinomial thus produced and write it as the square of a binomial x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 = (b/2a)2 – c/a apply the square root property to equate the new binomial to the constant on the right x + b/2a = (   b2 – 4ac )/2a subtract the constant term of the binomial from both sides to isolate x consolidate the constants on the right x = (– b   b2 – 4ac )/2a The result is the well known quadratic formula. Use of this formula will always produce a correct solution (or pair of solutions) – even when the solutions are not real numbers. x = b 2a 2 4ac + Quadratic Formula x = 2a b 2 4ac + 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Solving Quadratic Equations Example 2 Solve: 4x2 – 12x + 29 = 20 First put into standard form: 4x2 – 12x + 9 = 0 WHY ? Thus a = 4 , b = –12 , c = 9 x = 2a b2 – 4ac –b Apply Quadratic Formula The Quadratic Formula: Example 2 Here we start with an equation not in standard form, so the first step is to rewrite the equation in standard form. This allows us to identify the standard coefficients that appear in the quadratic formula. We note that the radicand is actually 0, so we get only one solution. Clearly there are not always two solutions. x = 2 4 ( ) –12 2 – 12 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Solving Quadratic Equations Example 2 Solve: 4x2 – 12x + 29 = 20 (continued) x = 2 4 ( ) –12 2 – 12 x = 8 144 – 144 12 = 8 12 x The Quadratic Formula: Example 2 Here we start with an equation not in standard form, so the first step is to rewrite the equation in standard form. This allows us to identify the standard coefficients that appear in the quadratic formula. We note that the radicand is actually 0, so we get only one solution. Clearly there are not always two solutions. { 2 3 } = 2 3 x Solution set: Question: Are there always two solutions ? 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Quadratic Formula Solving Quadratic Equations The Discriminant For ax2 + bx + c = 0 we found that The expression b2 – 4ac is called the discriminant It determines the number and type of solutions x = 2a b2 – 4ac –b The Discriminant We have noted that the value of the radicand, b2 – 4ac, controls how many solutions we will have for the quadratic equation. This value has a name: the discriminant. As shown, there are three possibilities for the discriminant value: b2 – 4ac = 0 yields exactly one real solution (since ) b2 – 4ac > 0 yields two real solutions, since every positive number has two real square roots b2 – 4ac < 0 yields two complex solutions, since negative numbers do not have any real square roots; the complex solutions occur in complex conjugate pairs: and These observations on the discriminant show that the quadratic formula helps us extend algebraic solutions beyond the real number system. However, we will confine our study to the real numbers, unless otherwise specified. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Quadratic Formula Solving Quadratic Equations The Discriminant Determines the number and type of solutions If b2 – 4ac = 0 have one real solution If b2 – 4ac > 0 have two real solutions If b2 – 4ac < 0 have two complex solutions The Discriminant We have noted that the value of the radicand, b2 – 4ac, controls how many solutions we will have for the quadratic equation. This value has a name: the discriminant. As shown, there are three possibilities for the discriminant value: b2 – 4ac = 0 yields exactly one real solution (since ) b2 – 4ac > 0 yields two real solutions, since every positive number has two real square roots b2 – 4ac < 0 yields two complex solutions, since negative numbers do not have any real square roots; the complex solutions occur in complex conjugate pairs: and These observations on the discriminant show that the quadratic formula helps us extend algebraic solutions beyond the real number system. However, we will confine our study to the real numbers, unless otherwise specified. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Factoring Solving Quadratic Equations Reduced Standard Form – Type 2 ax2 + bx = 0 , for a ≠ 0 Factor x and use zero-product property x(ax + b) = 0 Zero-product Property: if and only if p = 0 or q = 0 For any real numbers p and q , Quadratic Functions: Reduced Standard Form – Type 2 If the quadratic equation has no constant term, then we can write it as simply the product of x and a binomial in x equal to 0. This allows us to factor x from both terms and apply the zero-product property of the real numbers. Note that this property says at least one of the factors is zero, possibly both. Since we technically do not know which of these conditions holds, we assume that it might be both and solve two equations, generally yielding two solutions though it might be one solution, if the factors containing x are identical. The example illustrates a typical solution when the quadratic is of this reduced standard form – Type 2 (i.e. no constant term). pq = 0 ... or both 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Factoring Solving Quadratic Equations Reduced Standard Form – Type 2 ax2 + bx = 0 , for a ≠ 0 Example: Solve 3x2 – 5x = 0 x(3x – 5) = 0 From the zero-product property Quadratic Functions: Reduced Standard Form – Type 2 If the quadratic equation has no constant term, then we can write it as simply the product of x and a binomial in x equal to 0. This allows us to factor x from both terms and apply the zero-product property of the real numbers. Note that this property says at least one of the factors is zero, possibly both. Since we technically do not know which of these conditions holds, we assume that it might be both and solve two equations, generally yielding two solutions though it might be one solution, if the factors containing x are identical. The example illustrates a typical solution when the quadratic is of this reduced standard form – Type 2 (i.e. no constant term). x = 0 or 3x – 5 = 0 or both {0 , 5 3 } Solution Set: 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Special Forms Solving Quadratic Equations Conjugate Binomials Pairs of binomials of form (A + B) and (A – B) are called conjugate binomials Product of conjugate pair is always a difference of squares (A + B)(A – B) = A2 – B2 Special Forms: Conjugate Binomials To borrow a notion from the complex conjugate pairs mentioned earlier, we see that binomials A + B and A – B are real numbers that mimic their complex conjugate cousins. The reason we are interested in conjugate binomials is that their product is always the difference of squares of A and B, that is, (A + B)(A – B) = A2 – B2 . The examples show this fairly plainly. They also show that this relationship can be used to factor a difference of squares. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Special Forms Solving Quadratic Equations Conjugate Binomials Examples x2 – 4 9 – 4x2 4x2 – 9y2 = (x + 2)(x – 2) = (3 – 2x)(3 + 2x) = (2x – 3y)(2x + 3y) Special Forms: Conjugate Binomials To borrow a notion from the complex conjugate pairs mentioned earlier, we see that binomials A + B and A – B are real numbers that mimic their complex conjugate cousins. The reason we are interested in conjugate binomials is that their product is always the difference of squares of A and B, that is, (A + B)(A – B) = A2 – B2 . The examples show this fairly plainly. They also show that this relationship can be used to factor a difference of squares. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Forms Solving Quadratic Equations What if the expression is not a special form? We look for factors with integer coefficients In general these have form: x2 + (a + b)x + ab = (x + a)(x + b) We look for factors (x + a) and (x + b) that fit this form Quadratic Functions: Not-So-Special Forms This is an exercise in factoring quadratic expressions that are not one of the special forms. If we limit our focus to integer coefficients, then we have the general form of the product of two binomials as shown in the illustration. The general form shows what relationships we look for among the coefficients. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations What if the expression is not a special form? Factoring Examples: x2 + 7x + 12 (x + 3)(x + 4) = y2 + 13y + 40 (y + 5)(y + 8) = Quadratic Functions: Not-So-Special Forms This is an exercise in factoring quadratic expressions that are not one of the special forms. If we limit our focus to integer coefficients, then we have the general form of the product of two binomials as shown in the illustration. The general form shows what relationships we look for among the coefficients. 4x2 + 16x + 15 (2x + 3)(2x + 5) = Question: Is there a systematic way to do this ? 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Factoring Trinomials with negative constant term In general these have form: x2 + (a – b)x – ab = (x + a)(x – b) We look for factors that fit this form Not-So-Special Forms: One negative constant This form has a negative constant term with middle term either positive or negative. Analysis of the signs help us determine the signs and relative sizes of the constants in binomials we seek as factors in the trinomial. In the general form, assuming that a and b are themselves positive numbers, the product (a)(-b) must be negative. Since the last term in the trinomial is this product, the last term will always be negative when only one term in the binomials is negative. The middle term of the trinomial is the sum of the two products ax and -bx, which factors to (a – b)x. The sign of the factor (a – b) is determined by the relative sizes of a and b, to that the middle term in the trinomial could be either positive or negative. The examples clearly illustrate this point. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Examples x2 – x – 12 = y2 + 3y – 40 = 4x2 – 4x – 15 = Note the sign of last term: always negative Note the sign of the middle term: positive or negative (x + 3)(x – 4) (y – 5)(y + 8) (2x + 3)(2x – 5) WHY ? Not-So-Special Forms: One negative constant This form has a negative constant term with middle term either positive or negative. Analysis of the signs help us determine the signs and relative sizes of the constants in binomials we seek as factors in the trinomial. In the general form, assuming that a and b are themselves positive numbers, the product (a)(-b) must be negative. Since the last term in the trinomial is this product, the last term will always be negative when only one term in the binomials is negative. The middle term of the trinomial is the sum of the two products ax and -bx, which factors to (a – b)x. The sign of the factor (a – b) is determined by the relative sizes of a and b, to that the middle term in the trinomial could be either positive or negative. The examples clearly illustrate this point. WHY ? 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with two negative terms Sign of the larger term Terms a , b of opposite sign Factor x2 – 5x – 24 Find a and b such that x2 + (a + b)x + ab = (x + a)(x + b) Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Possible factors of –24 : 1, –24 2, –12 3, –8 4, –6 Since 3 – 8 = –5 , then a = 3 and b = –8 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with one negative term – the constant Sign of the larger term Terms a , b of opposite sign Factor x2 + 5x – 24 Find a and b such that x2 + (a + b)x + ab = (x + a)(x + b) Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Possible factors of –24 : –1, 24 –2, 12 –3, 8 –4, 6 Since 3 – 8 = 5 , then a = 8 and b = –3 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with one negative term – the constant Factor x2 + 5x – 24 Since 3 – 8 = 5 , then a = 8 and b = –3 This gives us: x2 + 5x – 24 = (x + 8)(x – 3) NOTE: Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. This is not a solution , since we are changing the form of an expression , NOT solving an equation 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with one negative term – the middle term Sign of both terms Terms a , b of same sign Solve x2 – 8x + 12 = 0 by factoring Find a and b such that x2 + (a + b)x + ab = (x + a)(x + b) Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Possible factors of 12 : –1, –12 –2, –6 –3, –4 Since –2 – 6 = –8 , then a = –2 and b = –6 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with one negative term – the middle term Solve x2 – 8x + 12 = 0 by factoring Since –2 – 6 = –8 , then a = –2 and b = –6 This gives us: x2 – 8x + 12 = (x – 2)(x – 6) = 0 So x – 2 = 0 or x – 6 = 0 Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Solution Set: 2, 6 { } NOTE: This is a solution , since we are solving an equation 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with no negative terms Sign of both terms Terms a , b of same sign Solve x2 + 11x + 28 = 0 by factoring Find a and b such that x2 + (a + b)x + ab = (x + a)(x + b) Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Possible factors of 28 : 1, 28 2, , 7 Since = 11 , then a = 4 and b = 7 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Not-So-Special Trinomials
Solving Quadratic Equations Trinomials with no negative terms Solve x2 + 11x + 28 = 0 by factoring Since = 11 , then a = 4 and b = 7 This gives us: x2 + 11x + 28 = (x + 4)(x + 7) = 0 So x + 4 = 0 or x + 7 = 0 Quadratic Functions: Not-So-Special Forms Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don’t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation – IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression – not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Solution Set: – 4 , –7 { } NOTE: This is a solution , since we are solving an equation 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Expanding a Room Solving Quadratic Equations A 9 by 12 room is to be expanded by the same amount x in length and width to double the area – what is x ? Original area: 12 + x A1 = 208 ft2 9 + x 12 ft 9 ft A1 = 108 x ft New area: A = 2A2 = 216 ft2 A2 = 9x Expanding A Room Here we apply our knowledge of the quadratic formula to the solution of a practical problem – how much to expand a room by to meet certain dimension conditions. We wish to double the area A1 of the room by expanding it x ft in both length and width. The new area A is thus 2A1. Since A1 = 108 then A = 2A1 = As the illustration indicates there are two ways to do this: by adding up the original area, together with the three new areas formed by the new rectangles, that is, A = A1 + A2 + A3 + A4 Or we can consider the new outside dimensions of the room 9 + x and x, and simply find the product of the two: A = (9 + x)(12 + x) which yields the same final equation to solve: x2 + 21x – 108 = 0 It doesn’t take long to figure out that this polynomial is not factorable, so we resort to using the quadratic formula. We discard the negative root as being not a real measure of length and so choose the positive root, approximately 4.3 feet. Two methods: A = (9 + x)(12 + x) x ft = 216 A3 = 12x A4 = x2 A = A1 + A2 + A3 + A4 = 216 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

Expanding a Room Solving Quadratic Equations Two methods: A = (9 + x)(12 + x) = 216 OR A = A1 + A2 + A3 + A4 = 216 Either way: x2 + 21x = 216 So x2 + 21x – 108 = 0 x = 2 1 ( ) 21 2 – 4 1 –108 21 x = 2 876 21 Expanding A Room Here we apply our knowledge of the quadratic formula to the solution of a practical problem – how much to expand a room by to meet certain dimension conditions. We wish to double the area A1 of the room by expanding it x ft in both length and width. The new area A is thus 2A1. Since A1 = 108 then A = 2A1 = As the illustration indicates there are two ways to do this: by adding up the original area, together with the three new areas formed by the new rectangles, that is, A = A1 + A2 + A3 + A4 Or we can consider the new outside dimensions of the room 9 + x and x, and simply find the product of the two: A = (9 + x)(12 + x) which yields the same final equation to solve: x2 + 21x – 108 = 0 It doesn’t take long to figure out that this polynomial is not factorable, so we resort to using the quadratic formula. We discard the negative root as being not a real measure of length and so choose the positive root, approximately 4.3 feet. ≈ 4.3 sq. ft. 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Quadratic Formula 1 Solving Quadratic Equations Complete the Square on the Standard Form ax2 + bx + c = 0 , for a ≠ 0 Rewrite: move c, divide by a c a b x2 x + = Now complete the square on the left side Want a constant k such that The Quadratic Formula We here solve the general quadratic equation in standard form by completing the square. The steps are exactly the same as before: divide out the lead coefficient to force the new lead coefficient to 1 subtract the constant term from both sides of the equation rewrite the linear term with a factor of 2 (multiply and divide by 2) x2 + 2(b/2a)x = -c/a square the resulting cofactor and add it to both sides of the equation x2 + 2(b/2a)x + (b/2a)2 = (b/2a)2 – c/a recognize the perfect-square trinomial thus produced and write it as the square of a binomial x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 = (b/2a)2 – c/a apply the square root property to equate the new binomial to the constant on the right x + b/2a = (   b2 – 4ac )/2a subtract the constant term of the binomial from both sides to isolate x consolidate the constants on the right x = (– b   b2 – 4ac )/2a The result is the well known quadratic formula. Use of this formula will always produce a correct solution (or pair of solutions) – even when the solutions are not real numbers. b a x2 x + + k2 is a perfect square – that is b a x2 x + + k2 + = x2 x b 2a ( ) 2 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

The Quadratic Formula 2 Solving Quadratic Equations Complete the Square on the Standard Form c a b x2 x + = b a x2 x + + k2 = 2a ( ) 2 k = b 2a = b 2a ( ) 2 k2 Thus …and + k2 b a x2 x c a = + k2 So … The Quadratic Formula We here solve the general quadratic equation in standard form by completing the square. The steps are exactly the same as before: divide out the lead coefficient to force the new lead coefficient to 1 subtract the constant term from both sides of the equation rewrite the linear term with a factor of 2 (multiply and divide by 2) x2 + 2(b/2a)x = -c/a square the resulting cofactor and add it to both sides of the equation x2 + 2(b/2a)x + (b/2a)2 = (b/2a)2 – c/a recognize the perfect-square trinomial thus produced and write it as the square of a binomial x2 + 2(b/2a)x + (b/2a)2 = (x + b/2a)2 = (b/2a)2 – c/a apply the square root property to equate the new binomial to the constant on the right x + b/2a = (   b2 – 4ac )/2a subtract the constant term of the binomial from both sides to isolate x consolidate the constants on the right x = (– b   b2 – 4ac )/2a The result is the well known quadratic formula. Use of this formula will always produce a correct solution (or pair of solutions) – even when the solutions are not real numbers. + x2 x b 2a ( ) 2 = b 2a ( ) 2 c a + x + ( b 2a ) 2 c a + b 2 4a = 7/9/2013 Quadratic Equations Quadratic Equations 7/9/2013

( ) Want k so that a = x2 + x + k2 (x + k)2 = x2 + 2kx + k2 , k2 = 2a
b = x x + k2 (x + k)2 = x2 + 2kx + k2 , k2 = 2a b ( ) 2 2k = a b , k = 2a b 7/9/2013 Quadratic Equations

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