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Lesson 6.6 , For use with pages 452-459

1. 64. 2. 3. 3. 5. 2. ( – 32). Lesson 6.6 , For use with pages 452-459. Evaluate the expression. . 16. ANSWER. –8. ANSWER. Lesson 6.6 , For use with pages 452-459. Evaluate the expression. . 3. Expand ( x – 5) 2. x 2 – 10 x + 25. ANSWER. Solve the equation. .

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Lesson 6.6 , For use with pages 452-459

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  1. 1. 64 2 3 3 5 2. (–32) Lesson 6.6, For use with pages 452-459 Evaluate the expression. 16 ANSWER –8 ANSWER

  2. Lesson 6.6, For use with pages 452-459 Evaluate the expression. 3.Expand (x– 5)2 x2– 10x + 25 ANSWER Solve the equation. 4.x2 + 6x + 9 = x + 45 ANSWER –9, 4

  3. Lesson 6.6, For use with pages 452-459 Solve the equation 5. A cone-shaped cup has a height of 7 cm and a radius of 2.5 cm. Use V =r2h to find the volume of the cup. 1 – π 3 about 45.8 cm3 ANSWER

  4. 6.6 Intermediate Algebra Solving Radical Equations

  5. 2x+7 2x+7 2x+7 Solve 3 = 3.    = 3 3 ( )3 3 = 33 2x+7 = 27 2x = 20 x = 10 EXAMPLE 1 Solve a radical equation Write original equation. Cube each side to eliminate the radical. Simplify. Subtract 7 from each side. Divide each side by 2.

  6. 3 27  2(10)+7 3 3 3 = 3 3 ? ? = = EXAMPLE 1 Solve a radical equation CHECK Check x = 10 in the original equation. Substitute 10 for x. Simplify. Solution checks.

  7. 2. ( x+25 ) = 4 3. (23 x –3 ) = 4 for Example 1 GUIDED PRACTICE Solve equation. Check your solution. 1.3√ x – 9 = –1 x = 512 ANSWER ANSWER x = 11 x = –9 ANSWER

  8. In a hurricane, the mean sustained wind velocity v(in meters per second) is given by v(p) = 6.3 1013 – p  where pis the air pressure (in millibars) at the center of the hurricane. Estimate the air pressure at the center of a hurricane when the mean sustained wind velocity is 54.5 meters per second. EXAMPLE 2 Solve a radical equation given a function Wind Velocity

  9. v(p) = 6.3 1013 – p  = 6.3 54.5 1013 – p 1013 – p 1013 – p    8.65 2 (8.65)2 74.8 1013 –p –p p 938.2 EXAMPLE 2 Solve a radical equation given a function SOLUTION Write given function. Substitute 54.5 for v(p). Divide each side by 6.3. Square each side. Simplify. – 938.2 Subtract 1013 from each side. Divide each side by –1.

  10. ANSWER The air pressure at the center of the hurricane is about 938millibars. EXAMPLE 2 Solve a radical equation given a function

  11. ANSWER about 954 mille bars. for Example 2 GUIDED PRACTICE 4. What If?Use the function in Example 2to estimate the air pressure at the center of a hurricane when the mean sustained wind velocity is 48.3meters per second.

  12. 3 Raise each side to the power . 2 EXAMPLE 3 Standardized Test Practice SOLUTION 4x2/3 = 36 Write original equation. x2/3 = 9 Divide each side by 4. (x2/3)3/2= 93/2 x = 27 Simplify. ANSWER The correct answer is D.

  13. 4/3 (x + 2)3/4 = 8 4/3 Raise each side to the power . 4 3 EXAMPLE 4 Solve an equation with a rational exponent Solve (x + 2)3/4 – 1 = 7. (x + 2)3/4 – 1 = 7 Write original equation. (x + 2)3/4 = 8 Add 1 to each side. x + 2 = (8 1/3)4 Apply properties of exponents. x + 2 = 24 Simplify. x + 2 = 16 Simplify. x = 14 Subtract 2 from each side.

  14. ANSWER The solution is 14. Check this in the original equation. EXAMPLE 4 Solve an equation with a rational exponent

  15. 7. = –2 x1/5 – 2 3 for Examples 3 and 4 GUIDED PRACTICE Solve the equation. Check your solution. 8. (x + 3)5/2 = 32 5. 3x3/2 = 375 ANSWER x = 25 ANSWER x = 1 6. –2x3/4 = –16 9. (x – 5)5/3 = 243 x = 32 ANSWER x = 16 ANSWER 10. (x + 2)2/3 +3 = 7 x = 243 ANSWER ANSWER x = –10 or 6

  16. Solve x + 1 = 7x + 15  x + 1 = 7x + 15  (x + 1)2 = ( 7x + 15)2  EXAMPLE 5 Solve an equation with an extraneous solution Write original equation. Square each side. Expand left side and simplify right side. x2 + 2x + 1 = 7x + 15 x2 – 5x – 14 = 0 Write in standard form. Factor. (x – 7)(x + 2) = 0 x – 7 = 0 or x + 2 = 0 Zero-product property Solve for x. x = 7 orx = –2

  17. Check x = 7 in the original equation. Check x = –2 in the original equation. x+ 1 = 7x+ 15 x+ 1 = 7x+ 15   7+ 1 –2+ 1 ?  = 7(–2) + 15 8 –1 = 7(7) + 15 ?  = 64 = 1 ? ?   8 = 8 –1 The only solution is 7. (The apparent solution –2 is extraneous.) ANSWER = 1 / EXAMPLE 5 Solve an equation with an extraneous solution CHECK

  18. Solve 2 2 METHOD 1 Solve using algebra. x + 2 + 1 = 3 – x   x + 2 + 1 x + 2 + 1   = 3 – x . = 3 – x   x + 2 +1 x + 2 +2   2 x + 2 EXAMPLE 6 Solve an equation with two radicals SOLUTION Write original equation. Square each side. = 3 – x Expand left side and simplify right side. = –2x Isolate radical expression.

  19. x + 2 x + 2   2 EXAMPLE 6 Solve an equation with two radicals = –x Divide each side by 2. = ( –x)2 Square each side again. = x2 x + 2 Simplify. 0 = x2 – x – 2 Write in standard form. = (x – 2)(x + 1) 0 Factor. x – 2 = 0 x + 1 = 0 or Zero-product property. x = –1 Solve for x. x = 2 or

  20. x+ 2 +1 x+ 2 +1   2+ 2 +1  –1+ 2 +1  ? = 3 – (–1)  ? = 3 – x = 3 – 2 = 3 – x    1 +1  = 1 ? = 4 ?    4 +1 2 = 2 3 ANSWER The only solution is 1. (The apparent solution 2 is extraneous.) = –1 / EXAMPLE 6 Solve an equation with two radicals Check x = 2 in the original equation. Check x = – 1 in the original equation.

  21. Use: a graph to solve the equation. Use a graphingcalculator to graphy1=+ 1andy2=. Thenfind the intersection points of the two graphs by using the intersect feature. You will find that the only point ofintersection is(–1, 2).Therefore,–1 is the onlysolution of the equation x + 2 + 1 = 3 – x EXAMPLE 6 Solve an equation with two radicals METHOD 2

  22. 12. 11.x – 1 13. 14. 10x + 9 = x + 3 = = x = 2 1 4 x + 6 – 2 2x + 5 x – 2 x + 7 for Examples 5 and 6 GUIDED PRACTICE Solve the equation. Check for extraneous solutions ANSWER ANSWER x = 9 x = 2 x = 0,4 ANSWER ANSWER x = 3

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