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Solving Quadratic Equations with Factoring and the Quadratic Formula

Learn how to solve quadratic equations by factoring and using the quadratic formula. Practice problems included.

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Solving Quadratic Equations with Factoring and the Quadratic Formula

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  1. Warm Up Hint: Grouping Hint: GCF first.. Then SUM of CUBES 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 ax2 + 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 Divide by 4 4t = – 1 t = – ¼ Add 5 3t – 5 = 0 3t = 5 Divide by 3 t = 5/3 Solution: t = - ¼ and 5/3  t = {- ¼, 5/3}

  5. Ex2: Solve x2 + 7x + 6 = 0 Quadratic equation  factor the left hand side (LHS) 6 1 x2 + 7x + 6 = (x + )(x + )  x2 + 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 + 1 = 0 x = – 1 x = – 6 Solution: x = - 6 and – 1  x = {-6, -1}

  6. Ex3: Solve x2 + 10x = – 25 Quadratic equation but not of the form ax2 + bx + c = 0  x2 + 10x + 25 = 0 Add 25 Quadratic equation  factor the left hand side (LHS) 5 x2 + 10x + 25 = (x + )(x + ) 5  x2 + 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 = 0 x = – 5 x = – 5 Solution: x = - 5  x = {- 5}  repeated root

  7. Ex4: Solve 12y2 – 5y = 2 Quadratic equation but not of the form ax2 + bx + c = 0  12y2 – 5y – 2 = 0 Subtract 2 Quadratic equation  factor the left hand side (LHS)

  8.  12y2 – 5y – 2 = 0  12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve 3y – 2 = 0 4y + 1 = 0 3y = 2 4y = – 1 y = 2/3 y = – ¼ Solution: y = 2/3 and – ¼  y = {2/3, - ¼ }

  9. Ex5: Solve 5x2 = 6x Quadratic equation but not of the form ax2 + bx + c = 0  5x2 – 6x = 0 Subtract 6x Quadratic equation  factor the left hand side (LHS) – 6 5x2 – 6x = x( ) 5x  5x2 – 6x = x(5x – 6) = 0 Now the equation as given is of the form ab = 0  set each factor equal to 0 and solve 5x – 6 = 0 x = 0 5x = 6 x = 6/5 Solution: x = 0 and 6/5  x = {0, 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 x2 = a, then x = +

  11. Ex6: Solve by taking square roots 3x2 – 36 = 0 3x2 – 36 = 0 First, isolate x2: 3x2 = 36 x2 = 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 Trinomial Binomial Square x2 + 8x + 16 (x + 4)2 x2– 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 x2 + 7x + 6 = 0 1 7 6 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in ax2 + bx + c = 0: a = b = c = 1 6 7 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 2m2 + m – 10 = 0 1 2 – 10 Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by Identify a, b, and c in am2 + bm + c = 0: a = b = c = 2 1 - 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. Homework Page 229 #9-17, 40, 57

  23. HW: (4-5) Page 229 #9-17, 40, 57 Mini Quiz Ch. 4 • GCF • Difference of Squares • Perfect Squares • Difference of Cubes • Grouping • Trinomials x2 + 8x + 16

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