EXAMPLE 2 Rationalize denominators of fractions. 5 2 3 7 + 2 Simplify

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Presentation transcript:

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

EXAMPLE 2 Rationalize denominators of fractions. SOLUTION = 3 7 + 2 7 – (b) 3 7 + 2 = 21 – 3 2 49 – 7 + 7 – 2 = 21 – 3 2 47

Solve a quadratic equation 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.

Solve a quadratic equation 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 – 3 + and – 3 – 35

EXAMPLE 4 Standardized Test Practice ANSWER The correct answer is C.

GUIDED PRACTICE GUIDED PRACTICE for Examples 2, 3, and 4 Simplify the expression. 6 5 SOLUTION 6 5 = 6 5 = 6 5 5 30 =

GUIDED PRACTICE GUIDED PRACTICE for Examples 2, 3, and 4 Simplify the expression. 9 8 SOLUTION 9 8 = 9 8 = 9 8 2 4 = 3

GUIDED PRACTICE GUIDED PRACTICE for Examples 2, 3, and 4 Simplify the expression. 17 12 SOLUTION 17 12 = 17 12 = 17 12 51 12 = 2 6

GUIDED PRACTICE GUIDED PRACTICE for Examples 2, 3, and 4 Simplify the expression. 19 21 SOLUTION 19 21 = 19 21 = 19 21 399 21 =

GUIDED PRACTICE for Examples 2, 3, and 4 – 6 7 – 5 SOLUTION = – 6 7 – 5 7 + – 6 7 – 5 = – 42 – 6 5 49 – 7 + 7 – 5 = – 21 – 3 5 22

GUIDED PRACTICE for Examples 2, 3, and 4 2 4 + 11 SOLUTION = 2 4 + 11 4 – 2 4 + 11 = 8 – 2 11 16 – 4 + 4 – 11 = 8 – 2 11 5

GUIDED PRACTICE for Examples 2, 3, and 4 – 1 9 + 7 SOLUTION – 1 9 + 7 = 9 – = – 9 + 7 81 – 9 + 9 – 7 = – 9 + 7 74

GUIDED PRACTICE for Examples 2, 3, and 4 4 8 – 3 SOLUTION = 4 8 – 3 8 + 4 8 – 3 = 32 + 4 3 64 – 4 + 4 – 3 = 32 + 4 3 61

GUIDED PRACTICE for Examples 2, 3, and 4 Solve the equation. 5x2 = 80 SOLUTION 5x2 = 80 Write original equation. x2 = 16 Divide each side by 5. x = + 16 Take square roots of each side. x = + 4 Product property x = + 4 Simplify.

GUIDED PRACTICE for Examples 2, 3, and 4 ANSWER The solutions are and . 4 – 4 Check Check the solutions by substituting them into the original equation. 5x2 = 80 5x2 = 80 5(4)2 = 80 ? 5(– 4)2 = 80 ? 5(16) = 80 ? 5(16) = 80 ? 80 = 80  80 = 80 

GUIDED PRACTICE for Examples 2, 3, and 4 Solve the equation. z2 – 7 = 29 SOLUTION z2 – 7 = 29 Write original equation. z2 = 36 Add 7 to each side. z = + 36 Take square roots of each side. z = + 6 Product property z = + 6 Simplify.

GUIDED PRACTICE for Examples 2, 3, and 4 ANSWER The solutions are and . 6 – 6 Check Check the solutions by substituting them into the original equation. z2 – 7 = 29 z2 – 7 = 29 ? (6)2 – 7 = 29 ? (– 6)2 – 7 = 29 ? 36 – 7 = 29 ? 36 – 7 = 29 29 = 29  29 = 29 

GUIDED PRACTICE for Examples 2, 3, and 4 3(x – 2)2 = 40 SOLUTION Write original equation. (x – 2)2 = 40 3 Divide each side by 3. + (x – 2) = 40 3 Take square roots of each side. x = 40 3 2 + Division property x = 40 3 2 + = 120 3 2 +

GUIDED PRACTICE for Examples 2, 3, and 4 = 4(30) 3 2 + = 2 30 3 2 + ANSWER The solutions are . 2 30 3 2 +