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Warm Up Hint: GCF first.. Then SUM of CUBES Hint: Grouping Hint: Diff of squares.

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Presentation on theme: "Warm Up Hint: GCF first.. Then SUM of CUBES Hint: Grouping Hint: Diff of squares."— Presentation transcript:

1 Warm Up Hint: GCF first.. Then SUM of CUBES Hint: Grouping Hint: Diff of squares

2 QUADRATIC EQUATIONS I <3 Los Al

3 Basics A quadratic equation is an equation equivalent to an equation of the type ax 2 + bx + c = 0, where a is nonzero We can solve a quadratic equation by factoring and using The Principle of Zero Products If ab = 0, then either a = 0, b = 0, or both a and b = 0.

4 Ex1: Solve (4t + 1)(3t – 5) = 0 Notice the equation as given is of the form ab = 0  set each factor equal to 0 and solve 4t + 1 = 0 Subtract 1 4t = – 1 Divide by 4 t = – ¼ 3t – 5 = 0 Add 5 3t = 5 Divide by 3 t = 5/3 Solution: t = - ¼ and 5/3  t = {- ¼, 5/3}

5 Ex2: Solve x 2 + 7x + 6 = 0 Quadratic equation  factor the left hand side (LHS) x 2 + 7x + 6 = (x + )(x + ) 61  x 2 + 7x + 6 = (x + 6)(x + 1) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve x + 6 = 0 x = – 6 x + 1 = 0 x = – 1 Solution: x = - 6 and – 1  x = {-6, -1}

6 Ex3: Solve x 2 + 10x = – 25 Quadratic equation but not of the form ax 2 + bx + c = 0 x 2 + 10x + 25 = (x + )(x + )5 5  x 2 + 10x + 25 = (x + 5)(x + 5) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve x + 5 = 0 x = – 5 x + 5 = 0 x = – 5 Solution: x = - 5  x = {- 5}  repeated root Quadratic equation  factor the left hand side (LHS) Add 25  x 2 + 10x + 25 = 0

7 Ex4: Solve 12y 2 – 5y = 2 Quadratic equation but not of the form ax 2 + bx + c = 0 Quadratic equation  factor the left hand side (LHS) Subtract 2  12y 2 – 5y – 2 = 0

8 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve 3y – 2 = 0 3y = 2 4y + 1 = 0 4y = – 1 Solution: y = 2/3 and – ¼  y = {2/3, - ¼ } y = 2/3y = – ¼  12y 2 – 5y – 2 = (3y - 2)(4y + 1) = 0

9 Ex5: Solve 5x 2 = 6x Quadratic equation but not of the form ax 2 + bx + c = 0 5x 2 – 6x = x( )5x – 6  5x 2 – 6x = x(5x – 6) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve x = 0 5x – 6 = 0 5x = 6 Solution: x = 0 and 6/5  x = {0, 6/5} Quadratic equation  factor the left hand side (LHS) Subtract 6x  5x 2 – 6x = 0 x = 6/5

10 Solving by taking square roots An alternate method of solving a quadratic equation is using the Principle of Taking the Square Root of Each Side of an Equation If x 2 = a, then x = +

11 Ex6: Solve by taking square roots 3x 2 – 36 = 0 First, isolate x 2 : 3x 2 – 36 = 0 3x 2 = 36 x 2 = 12 Now take the square root of both sides:

12 Ex7: Solve by taking square roots 4(z – 3) 2 = 100 First, isolate the squared factor: 4(z – 3) 2 = 100 (z – 3) 2 = 25 Now take the square root of both sides: z – 3 = + 5 z = 3 + 5  z = 3 + 5 = 8 and z = 3 – 5 = – 2

13 Ex8: Solve by taking square roots 5(x + 5) 2 – 75 = 0 First, isolate the squared factor: 5(x + 5) 2 = 75 (x + 5) 2 = 15 Now take the square root of both sides:

14 Perfect Square Trinomials Recall from factoring that a Perfect-Square Trinomial is the square of a binomial: Perfect square TrinomialBinomial Square x 2 + 8x + 16(x + 4) 2 x 2 – 6x + 9(x – 3) 2 The square of half of the coefficient of x equals the constant term: ( ½ * 8 ) 2 = 16 [½ (-6)] 2 = 9

15 Remember if you can’t factor, Use the Quadratic formula The only problems that don’t work are equations where the parabola does not cross the x-axis.

16 Ex9: Use the Quadratic Formula to solve x 2 + 7x + 6 = 0 Recall: For quadratic equation ax 2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in ax 2 + bx + c = 0: a = b = c =1 1 7 7 6 6 Now evaluate the quadratic formula at the identified values of a, b, and c

17 x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6 x = { - 1, - 6 }

18 Ex10: Use the Quadratic Formula to solve 2m 2 + m – 10 = 0 Recall: For quadratic equation ax 2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in am 2 + bm + c = 0: a = b = c =2 2 1 1 - 10 – 10 Now evaluate the quadratic formula at the identified values of a, b, and c

19 m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2 m = { 2, - 5/2 }

20 Any questions...

21 You can check your own work If you factor a problem and then set the factors equal to zero. You will get your x intercepts. You can also get your x intercepts by solving the quadratic equation. So you can check your problems by solving using both methods. If your answers match. You are correct!!

22 Page 229 #9-17, 40, 57

23 Mini Quiz Ch. 4 1)GCF 2)Difference of Squares 3)Perfect Squares 4)Difference of Cubes 5)Grouping 6)Trinomials x 2 + 8x + 16 HW: (4-5) Page 229 #9-17, 40, 57


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