1. Section 6.6 Solving Quadratic Equations Math in Our World

2. Learning Objectives  Identify the standard form of a quadratic equation.  Multiply binomials using FOIL.  Factor trinomials.  Solve quadratic equations using factoring.  Solve quadratic equations using the quadratic formula.  Solve real-world problems using quadratic equations.

3. Quadratic Equations A quadratic equation is any equation that can be written in the form ax 2 + bx + c = 0, where a, b, and c are real numbers, and a is not zero. When written this way, we say that a quadratic equation is in standard form. To write a quadratic equation in standard form, we arrange it so that zero is on the right side of the equation, the term with exponent 2 comes first on the left side, followed by the term with exponent 1 (if there is one) and then the constant (numeric) term.

4. EXAMPLE 1 Writing a Quadratic Equation in Standard Form Write each equation in standard form and identify a, b, and c. (a) 7 + 9x 2 = 3x (b) 4x – 15 = 3x 2 (c) 5x 2 = 25

5. EXAMPLE 1 Writing a Quadratic Equation in Standard Form SOLUTION (a) Now we use the definition of standard form: a is the coefficient of x 2, b is the coefficient of x, and c is the constant term, so we get a = 9, b = – 3, and c = 7. (b) Notice that if we wanted to, we could multiply both sides of the equation in standard form by – 1, giving us the equivalent equation 3x 2 – 4x + 15 = 0. In this case, a = 3, b = – 4, and c = 15.

6. EXAMPLE 1 Writing a Quadratic Equation in Standard Form SOLUTION (c)

7. Binomials A binomial is an algebraic expression with two terms in which any variable has a whole number exponent. Some examples of binomials are

8. Multiplying Binomials FOIL Method: F represents the product of the first terms of the binomial. O represents the product of the outer terms. I represents the product of the inner terms. L represents the product of the last terms of the binomials.

9. EXAMPLE 2 Multiplying Binomials using FOIL Multiply (x – 8)(x + 3).

10. EXAMPLE 2 Multiplying Binomials using FOIL Multiply the first terms: x · x = x 2 Multiply the outer terms: x · 3 = 3x Multiply the inner terms: – 8 · x = – 8x Multiply the last terms: (– 8) · (3) = – 24 Combining like terms, the product is x 2 – 5x – 24. SOLUTION

11. EXAMPLE 3 Multiplying Binomials using FOIL Multiply (2x – 5)(3x – 8). SOLUTION

12. Factoring Trinomials When a quadratic equation is written in standard form, there are three terms on the left side if none of the coefficients are zero. Three term expressions like this are called trinomials. It’s important that you’re really good at using FOIL, because one of our methods for solving quadratic equations will involve performing FOIL in reverse, a process we call factoring.

13. Factoring Trinomials Take another look at the result of Example 2: If we read this equation from right to left, it’s an example of factoring, because we start with the trinomial x 2 – 5x – 24 and write it as a product of two factors: x – 8 and x + 3. The x 2 comes from multiplying the first terms of the two factors, x and x. The constant term, 24, comes from multiplying the last two numbers of the two factors, 8 and 3. And the x term comes from combining the products of the outsides and insides.

14. EXAMPLE 4 Factoring a Trinomial with a = 1 Factor x 2 + 10x + 16. SOLUTION Step 1 The factored form will be the product of two binomials; the first term of each is x (since the product of the firsts must be x 2 ). So x 2 + 10x + 16 = (x )(x ) Step 2 Write all pairs of factors of the constant term. One of these pairs must be the product of the last terms inside the parentheses. In this case, they are 1 16, 2 8, and 4 4.

15. EXAMPLE 4 Factoring a Trinomial with a = 1 SOLUTION Step 3 Set up the factorization with each possible pair of last terms, found in Step 2. Then find the product of outside and inside terms for each possibility. 1 16: (x 1)(x 16) Outsides: 16x Insides: x 2 8: (x 2)(x 8) Outsides: 8x Insides: 2x 4 4: (x 4)(x 4) Outsides: 4x Insides: 4x Step 4 Find a sum or difference of the outsides and insides that yields the x term in the trinomial. In this case, we need 10x, and 8x + 2x = 10x. That tells us that the two last terms we need are +8 and +2, and the trinomial factors as x 2 + 10x + 16 = (x + 8)(x + 2)

16. EXAMPLE 5 Factoring a Trinomial with a = 1 Factor x 2 – 7x + 12. SOLUTION Step 1 Write as a product of two binomials with first term x. x 2 – 7x + 12 = (x )(x ) Step 2 The pairs of factors of the constant term are 1 12, 2 6, and 3 4.

17. EXAMPLE 5 Factoring a Trinomial with a = 1 SOLUTION Step 3 Possible factorizations. (x 1)(x 12) Outsides: 12x Insides: x (x 2)(x 6) Outsides: 6x Insides: 2x (x 3)(x 4) Outsides: 3x Insides: 4x Step 4 The pair of outsides and insides that yields middle term – 7x is 4x and 3x, if both are negative: – 4x – 3x = – 7x. Also, (– 4)(– 3) = 12, which is the correct constant term. So the factors are x 2 – 7x + 12 = (x – 3)(x – 4)

18. EXAMPLE 6 Factoring a Trinomial with a = 1 Factor x 2 – 2x – 8. SOLUTION Step 1 Write as a product of two binomials with first term x. x 2 – 2x – 8 = (x )(x ) Step 2 The pairs of factors of the constant term are 1 8 and 2 4.

19. EXAMPLE 6 Factoring a Trinomial with a = 1 SOLUTION Step 3 Possible factorizations. (x 1)(x 8) Outsides: 8x Insides: x (x 2)(x 4) Outsides: 4x Insides: 2x Step 4 The pair of outsides and insides that yields middle term – 2x is – 4x and 2x; – 4x + 2x = – 2x. Also, (– 4)(2) = – 8, which is the correct constant term. So the factors are x 2 – 2x – 8 = (x + 2)(x – 4)

20. EXAMPLE 7 Factoring a Trinomial with a ≠ 1 Factor 3x 2 – 4x – 15. SOLUTION Step 1 Write as a product of two binomials; the product of the first two terms has to be 3x 2, so we will use 3x and x. 3x 2 – 4x – 15 = (3x )(x ) Step 2 The pairs of factors of the constant term are 1 15 and 3 5.

21. EXAMPLE 7 Factoring a Trinomial with a ≠ 1 SOLUTION Step 3 Possible factorizations (important note: this time the order matters, so we have to write the possible combinations in both orders): (3x 1)(x 15) Outsides: 45x Insides: x (3x 15)(x 1) Outsides: 3x Insides: 15x (3x 3)(x 5) Outsides: 15x Insides: 3x (3x 5)(x 3) Outsides: 9x Insides: 5x Step 4 The pair of outsides and insides that yields middle term – 4x is – 9x and 5x; – 9x + 5x = – 4x. This comes from the factor 5 and –3. So the factors are 3x 2 – 4x – 15 = (3x + 5)(x – 3)

22. EXAMPLE 8 Factoring a Trinomial with a ≠ 1 Factor 14x 2 – 33x + 10. SOLUTION Step 1 Write as a product of two binomials; the product of the first two terms has to be 14x 2, so we can use 14x and x or 7x and 2x. 14x 2 – 33x + 10 = (14x )(x ) or (7x )(2x ) Step 2 The pairs of factors of the constant term are 1 10 and 2 5.

23. EXAMPLE 8 Factoring a Trinomial with a ≠ 1 SOLUTION Step 3 We’ll start trying the possible combinations starting with 14x and x. (14x 10)(x 1) Outsides: 14x Insides: 10x (14x 1)(x 10) Outsides: 140x Insides: x (14x 5)(x 2) Outsides: 28x Insides: 5x Step 4 We haven’t written all the combinations, but at this point we can stop. With factors (14x – 5) and (x – 2), we’ll get outsides – 28x and insides – 5x, which gives us the – 33x we need. 14x 2 – 33x + 10 = (14x – 5)(x – 2)

24. Solving Quadratic Equations Solving Quadratic Equations by Factoring Step 1 Write the quadratic equation in standard form. Step 2 Factor the left side. Step 3 Set both factors equal to zero. Step 4 Solve each equation for x. This works since the only way a product can equal zero is if one or both of the factors are equal to zero.

25. EXAMPLE 9 Solving a Quadratic Equation Using Factoring Solve x 2 – 13x = – 36. SOLUTION The solution set is {9, 4}. They can be checked by substituting the values into the original equation.

26. EXAMPLE 10 Solving a Quadratic Equation Using Factoring Solve 6x 2 – 6 = – 5x. SOLUTION

27. Solving Quadratic Equations Solving Using the Quadratic Formula Step 1 Write the equation in standard form. Step 2 Identify a, b, and c. Step 3 Substitute the values into the formula. If ax 2 + bx + c = 0, then

28. EXAMPLE 11 Using the Quadratic Formula Solve 2x 2 – x – 8 = 0 using the quadratic formula. SOLUTION The equation is in standard form, so a = 2, b = –1 and c = –8.

29. EXAMPLE 11 Using the Quadratic Formula SOLUTION Substitute a = 2, b = –1 and c = –8 into the formula.

30. EXAMPLE 12 Using the Quadratic Formula Solve 5 = 5y 2 + 8y using the quadratic formula. SOLUTION Before identifying a, b, and c, we need to rewrite the equation in standard form by subtracting 5y 2 and 8y from both sides. – 5y 2 – 8y + 5 = 0 So a = – 5, b = –8 and c = 5.

31. EXAMPLE 12 Using the Quadratic Formula SOLUTION Substitute a = – 5, b = –8 and c = 5 into the formula.

32. EXAMPLE 13 Applying Quadratic Equations to Woodworking The plans for a make-it-yourself picnic table call for the length to be 2 feet more than the width. If you want a table with 15 square feet of area, what dimensions should you choose? SOLUTION Let x = the width of the table. The length is 2 feet more, so x + 2 is the length. The area of a rectangle is length times width, so we can set up an equation: Length times Width is Area (x + 2) x = 15

33. EXAMPLE 13 Applying Quadratic Equations to Woodworking SOLUTION Now we solve. Even though we got two solutions, only one makes sense in the problem – you can’t build a table that is – 5 feet wide. So the width is 3 feet and the length is 5 feet.