 ## Presentation on theme: "3.6 Solving Quadratic Equations"— Presentation transcript:

By Finding the Square Root

EXAMPLE 1 Use properties of square roots Simplify the expression. a. 80 5 16 = 5 = 4 b. 6 21 126 = 9 14 = = 3 14 c. 4 81 = 4 81 = 2 9 d. 7 16 = 7 16 = 4 7

GUIDED PRACTICE GUIDED PRACTICE 27 10 15 9 64 ANSWER ANSWER ANSWER 3 8 3 6 5 8 28 15 4 98 ANSWER 2 15 ANSWER ANSWER 14 4 2 7 36 49 11 25 7 6 5 11

EXAMPLE 2 Rationalize denominators of fractions. 5 2 3 7 + 2 Simplify (a) and (b) SOLUTION = 3 7 + 2 7 – (a) 5 2 = 5 2 (b) 3 7 + 2 = 5 2 = 21 – 3 2 49 – – 2 2 10 = = 21 – 3 2 47

Key to solving using square roots
Remember—You have to include the positive (principle) root as well as the negative root. x2 = 121 x = ± 11

EXAMPLE 3 Solve a quadratic equation Solve 3x2 + 5 = 41. 3x2 + 5 = 41 Write original equation. 3x2 = 36 Subtract 5 from each side. x2 = 12 Divide each side by 3. x = + 12 Take square roots of each side. x = + 4 3 Product property x = + 2 3 Simplify.

EXAMPLE 3 Solve a quadratic equation ANSWER The solutions are and 2 3 2 3 Check the solutions by substituting them into the original equation. 3x2 + 5 = 41 3x2 + 5 = 41 3( )2 + 5 = 41 2 3 ? 3( )2 + 5 = 41 – 2 3 ? 3(12) + 5 = 41 ? 3(12) + 5 = 41 ? 41 = 41 41 = 41

Standardized Test Practice
EXAMPLE 4 Standardized Test Practice SOLUTION 15 (z + 3)2 = 7 Write original equation. (z + 3)2 = 35 Multiply each side by 5. z + 3 = + 35 Take square roots of each side. z = –3 + 35 Subtract 3 from each side. The solutions are – and –3 – 35

GUIDED PRACTICE GUIDED PRACTICE Simplify the expression. 6 5 19 21 5 30 399 21 ANSWER ANSWER 9 8 – 6 7 – 5 2 4 3 – 21 – 3 5 22 ANSWER ANSWER 17 12 2 4 + 11 51 6 ANSWER 8 – 2 11 5 ANSWER

GUIDED PRACTICE – 1 9 + 7 – 9 + 7 74 ANSWER 4 8 – 3 32 + 4 3 61 ANSWER

GUIDED PRACTICE Solve the equation. 5x2 = 80 ANSWER + 4 z2 – 7 = 29 + 6 ANSWER 3(x – 2)2 = 40 120 3 2 + ANSWER