Unit 5 Solving Quadratics By Square Roots Method and Completing the Square

Objectives I can solve a quadratic equation using the Square Roots Method I can solve a quadratic equation using the Complete the Square Method

Big Concept Anytime you take the Square Root of a variable when solving an equation you get the positive and negative answer

Square Roots Method Get the squared term on the left Get the number on the right Remember when you take the square root you get two answers (positive and negative)

Undoing a Squared Term To undo a squared term, take the square root Use all other normal algebra skills to solve an equation

Example 1

Example 2

Example 3

Example 4

Example 5

A Perfect Square A perfect square is a trinomial expression that has 2 factors that are the same: Example: (x x + 25) is a perfect square with factors (x + 5)(x + 5)

Special Factoring x x + 81 = 0 (x + 9)(x + 9) = 0 (x + 9) 2 = 0 This was a perfect square x 2 – 8x + 16 = 0 (x –4)(x – 4) = 0 (x – 4) 2 = 0 Again, a perfect square

Making a Perfect Square Consider the following equation: x x + c = 0 What number does c need to be to make a perfect square? Follow the procedure on next slide.

Perfect Square Method x x + c Take middle term and divide by 2 14/2 = 7 Next square the results 7 2 = 49 x x + 49 (x + 7)(x + 7) (x + 7) 2

Another Example x 2 – 8x + c Take middle term and divide by 2 -8/2 = - 4 Now square that new number (-4) 2 = 16 x 2 – 8x + 16 (x – 4) 2

You can get fractions x 2 – 5x + c Take middle term and divide by 2 -5/2 = - 5/2 Now square that new number (-5/2) 2 = 25/4 x 2 – 8x + 25/4 (x – 5/2) 2

Teeter-Toter Keeping Balanced +20

Solving by Completing the Square Given a quadratic equation ax 2 + bx + c = 0 Step 1: Move number to Right Side of Equation if necessary Step 2: Make the Left side a Perfect Square Step 3: Solve using Square Roots Method learned earlier this unit

Solving by Taking Square Root x 2 – 6x + 9 = 25 (x –3)(x – 3) = 25 (x - 3) 2 = 25 x – 3 = +/- 5 x = 3 +/- 5 x = 8 or –2 {-2, 8}

Teeter-Toter Keeping Balanced +20

Example 1 x 2 – 6x – 40 = 0 x 2 – 6x = 40 x 2 – 6x + _____ = 40 + _____ x 2 – 6x + 9 = (x – 3) 2 = 49 x – 3 =  x = 3  7 x = 10 or -4

Example 2 x 2 + 8x + 20 = 0 x 2 + 8x = -20 x 2 + 8x + ____ = ____ x 2 + 8x + 16 = (x + 4) 2 = -4 x + 4 =  x = -4  2i

GUIDED PRACTICE for Examples 3, 4 and 5 Solve x 2 + 6x + 4 = 0 by completing the square. x 2 + 6x + 4 = 0 Write original equation. x 2 + 6x = – 4 Write left side in the form x 2 + bx. x 2 + 6x + 9 = – Add ( ) = (3) 2 = 9 to each side. (x + 3) 2 = 5 Write left side as a binomial squared. Solve for x. Take square roots of each side. x + 3 = + 5 x = – The solutions are – 3+ and – 3 – 2 5 ANSWER 7.

GUIDED PRACTICE for Examples 3, 4 and 5 Solve 3x x – 18 = 0 by completing the square. Write original equation. x 2 + 4x = 6 Write left side in the form x 2 + bx. x 2 + 4x + 4 = Add ( ) = (2) 2 = 4 to each side. (x + 2) 2 = 10 Write left side as a binomial squared. Solve for x. Take square roots of each side. x + 2 = + 10 x = – x x – 18 = 0 Divided each side by the coefficient of x 2. x 2 + 4x – 6 = 0

Homework WS 5-4