Algebra 2 Perfect Squares Lesson 4 6 Part

Algebra 2 Perfect Squares Lesson 4 -6 Part 1

Goals Goal • To solve quadratic equations by finding square roots. • To solve a perfect square trinomial equation. Rubric Level 1 – Know the goals. Level 2 – Fully understand the goals. Level 3 – Use the goals to solve simple problems. Level 4 – Use the goals to solve more advanced problems. Level 5 – Adapts and applies the goals to different and more complex problems.

Vocabulary • Square Root Property

Essential Question Big Idea: Solving Equations • Why is a perfect square trinomial useful?

Perfect Square Quadratics Many quadratic equations contain expressions that cannot be easily factored. For equations containing these types of expressions, you can use square roots to find roots. Quadratic equations that do not contain a linear term (x – term) and quadratic equations that are perfect square trinomials can be solved using the square property.

Square Root Property Reading Math Read as “plus or minus square root of a. ”

Square Root Property If x 2 = k, then

Square-Root Property That is, the solution of If k = 0, then this is sometimes called a double solution.

Procedure Solving Quadratic Equations Using the Square Root Property Step 1: Isolate the expression containing the square term. Step 2: Use the Square Root Method. Don’t forget the symbol. Step 3: Isolate the variable, if necessary. Step 4: Check. Verify your solutions.

Example: Solve each quadratic equation. Solution: By the square root property, the solution is 1. 4 - 10

Example: Solve the quadratic equation. Solution: Use a generalization of the square root property. Generalized square root property. Add 4. 1. 4 - 11

Example: Solve the equation. 4 x 2 + 1 = 65 4 x 2 = 64 x 2 = 16 Subtract 1 from both sides. Divide both sides by 4 to isolate the square term. Take the square root of both sides. Simplify.

Example: Solve x 2 + 14 x + 49 = 64 by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Subtract 7 from each side.

Continued x = – 7 + 8 or x = – 7 – 8 x=1 x = – 15 Write as two equations. Solve each equation. Answer: The solution is x = – 15, 1. Check: Substitute both values into the original equation. x 2 + 14 x + 49 = 64 2 ? 1 + 14(1) + 49 = 64 ? 1 + 14 + 49 = 64 64 = 64 x 2 + 14 x + 49 = 64 2 ? (– 15) + 14(– 15) + 49 = 64 ? 225 + (– 210) + 49 = 64 64 = 64

Example: Solve the equation. x 2 + 12 x + 36 = 28 (x + 6)2 = 28 Factor the perfect square trinomial Take the square root of both sides. Subtract 6 from both sides. Simplify.

Your Turn: Solve the equation. 4 x 2 – 20 = 5 4 x 2 = 25 Add 20 to both sides. Divide both sides by 4 to isolate the square term. Take the square root of both sides. Simplify.

Your Turn: Solve the equation. x 2 + 8 x + 16 = 49 (x + 4)2 = 49 Factor the perfect square trinomial. Take the square root of both sides. x = – 4 ± x = – 11, 3 Subtract 4 from both sides. Simplify.

Your Turn: Solve x 2 – 4 x + 4 = 13 by using the Square Root Property. Original equation Factor the perfect square trinomial. Square Root Property Add 2 to each side. Write as two equations. Use a calculator.

Continued Answer: The exact solutions of this equation are The approximate solutions are 5. 61 and – 1. 61.

Your Turn: Solve x 2 – 16 x + 64 = 25 by using the Square Root Property. A. {– 1, 9} B. {11, 21} C. {3, 13} D. {– 13, – 3}

Your Turn: Solve x 2 – 4 x + 4 = 8 by using the Square Root Property. A. B. C. D.

Essential Question Big Idea: Solving Equations • Why is a perfect square trinomial useful? • Factor the perfect square trinomial as the square of a binomial. Then you can solve the quadratic equation by finding square roots.

Assignment • Section 4 -6 Part 1, Pg 254; # 4 -15