1. QUADRATIC EQUATIONS MSJC ~ San Jacinto Campus Math Center Workshop Series Janice Levasseur

2. 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.

3. Ex: 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}

4. Ex: 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}

5. Ex: 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

6. Ex: Solve 12y 2 – 5y = 2 Quadratic equation but not of the form ax 2 + bx + c = 0 ac method  a = 12 and c = – 2 = (3y – 2)(4y + 1) Quadratic equation  factor the left hand side (LHS) Subtract 2  12y 2 – 5y – 2 = 0 ac = (12)(-2) = - 24  factors of – 24 that sum to - 5 1&-24, 2&-12, 3&-8,...  12y 2 – 5y – 2 = 12y 2 + 3y – 8y – 2 = 3y(4y + 1) – 2(4y + 1)

7.  12y 2 – 5y – 2 = 0 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

8. Ex: 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

9. 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 = +

10. Ex: 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:

11. Ex: 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

12. Ex: 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:

13. Completing the Square 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

14. Completing the Square Write the equation in the form x 2 + bx = c Add to each side of the equation [½(b)] 2 Factor the perfect-square trinomial x 2 + bx + [½(b)] 2 = c + [½(b)] 2 Take the square root of both sides of the equation Solve for x

15. Ex: Solve w 2 + 6w + 4 = 0 by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. w 2 + 6w = – 4 Add [½(b)] 2 to both sides:b = 6 6  [½(6)] 2 = 3 2 = 9 w 2 + 6w + 9 = – 4 + 9 w 2 + 6w + 9 = 5 (w + 3) 2 = 5 Now take the square root of both sides

17. Ex: Solve 2r 2 = 3 – 5r by completing the square First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1. 2r 2 + 5r = 3 Add [½(b)] 2 to both sides:b = (5/2) 5/2  [½( 5/2 )] 2 = (5/4) 2 = 25/16 r 2 + (5/2)r + 25/16 = (3/2) + 25/16 r 2 + (5/2)r + 25/16 = 24/16 + 25/16 (r + 5/4) 2 = 49/16 Now take the square root of both sides  r 2 + (5/2)r = (3/2)

18. r = - (5/4) + (7/4) = 2/4 = ½ and r = - (5/4) - (7/4) = -12/4 = - 3 r = { ½, - 3}

19. Ex: Solve 3p – 5 = (p – 1)(p – 2) Is this a quadratic equation? FOIL the RHS 3p – 5 = p 2 – 2p – p + 2 3p – 5 = p 2 – 3p + 2 p 2 – 6p + 7 = 0 Collect all terms A-ha... Quadratic Equation  complete the square p 2 – 6p = – 7  [½(-6)] 2 = (-3) 2 = 9 p 2 – 6p + 9 = – 7 + 9 (p – 3) 2 = 2

21. The Quadratic Formula Consider a quadratic equation of the form ax 2 + bx + c = 0 for a nonzero Completing the square

22. The Quadratic Formula Solutions to ax 2 + bx + c = 0 for a nonzero are

23. Ex: 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

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

25. Ex: 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

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

27. Any questions...