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Solve radical equations

Objective. Solve radical equations. A radical equation contains a variable within a radical. Recall that you can solve quadratic equations by taking the square root of both sides. Similarly, radical equations can be solved by raising both sides to a power.

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Solve radical equations

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  1. Objective Solve radical equations

  2. A radical equationcontains a variable within a radical. Recall that you can solve quadratic equations by taking the square root of both sides. Similarly, radical equations can be solved by raising both sides to a power.

  3. The nth root of a real number a can be written as the radical expression , where n is the index (plural: indices) of the radical and a is the radicand. When a number has more than one root, the radical sign indicates only the principal, or positive, root.

  4. Remember! For a square root, the index of the radical is 2.

  5. Example 1A: Solving Equations Containing One Radical Solve each equation. Check Subtract 5. Simplify. Square both sides.  Simplify. Solve for x.

  6. 7 5x - 7 84 3 = 7 7 7 Example 1B: Solving Equations Containing One Radical Solve each equation. Check Divide by 7. Simplify. Cube both sides.  Simplify. Solve for x.

  7. Check It Out! Example 1a Solve the equation. Check Subtract 4. Simplify.  Square both sides. Simplify. Solve for x.

  8. Check It Out! Example 1b Solve the equation. Check Cube both sides. Simplify. Solve for x. 

  9. Check It Out! Example 1c Solve the equation. Check Divide by 6. Square both sides. Simplify.  Solve for x.

  10. Solve Example 2: Solving Equations Containing Two Radicals Square both sides. 7x + 2 = 9(3x – 2) Simplify. 7x + 2 = 27x – 18 Distribute. 20 = 20x Solve for x. 1 = x

  11. Example 2 Continued Check  3 3

  12. Check It Out! Example 2a Solve each equation. Square both sides. 8x + 6 = 9x Simplify. Solve for x. 6 = x Check 

  13. Check It Out! Example 2b Solve each equation. Cube both sides. x + 6 = 8(x – 1) Simplify. x + 6 = 8x – 8 Distribute. 14 = 7x Solve for x. 2 = x Check  2 2

  14. Helpful Hint You can use the intersect feature on a graphing calculator to find the point where the two curves intersect. Raising each side of an equation to an even power may introduce extraneous solutions.

  15. Example 3 Continued Method 1 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. –3x + 33 = 25 – 10x + x2 0 = x2 – 7x – 8 Write in standard form. 0 = (x – 8)(x + 1) Factor. x – 8 = 0 or x + 1 = 0 Solve for x. x = 8 or x = –1

  16. Example 3 Continued Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. 3 –3 6 6  x Because x = 8 is extraneous, the only solution is x = –1.

  17. Let Y1 = and Y2 = 5 – x. Example 3: Solving Equations with Extraneous Solutions Solve . Method 2 Use a graphing calculator. The graphs intersect in only one point, so there is exactly one solution. The solution is x = – 1

  18. Check It Out! Example 3a Continued Method 1 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. 2x + 14 = x2 + 6x + 9 Write in standard form. 0 = x2 + 4x – 5 Factor. 0 = (x + 5)(x – 1) x + 5 = 0 or x – 1 = 0 Solve for x. x = –5 or x = 1

  19. Check It Out! Example 3a Continued Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions. x 2 –2 4 4  Because x = –5 is extraneous, the only solution is x = 1.

  20. Let Y1 = and Y2 = x +3. Check It Out! Example 3a Solve each equation. Method 2 Use a graphing calculator. The graphs intersect in only one point, so there is exactly one solution. The solution is x = 1.

  21. Let Y1 = and Y2 = –x +4. Check It Out! Example 3b Solve each equation. Method 1 Use a graphing calculator. The graphs intersect in two points, so there are two solutions. The solutions are x = –4 and x = 3.

  22. Check It Out! Example 3b Continued Method 2 Use algebra to solve the equation. Step 1 Solve for x. Square both sides. Simplify. –9x + 28 = x2 – 8x + 16 Write in standard form. 0 = x2 + x – 12 Factor. 0 = (x + 4)(x – 3) x + 4 = 0 or x – 3 = 0 Solve for x. x = –4 or x = 3

  23. Check It Out! Example 3b Continued Method 1 Use algebra to solve the equation. Step 2 Use substitution to check for extraneous solutions.  

  24. Remember! To find a power, multiply the exponents.

  25. 1 3 Example 4A: Solving Equations with Rational Exponents Solve each equation. (5x + 7) = 3 Write in radical form. Cube both sides. 5x + 7 = 27 Simplify. 5x = 20 Factor. Solve for x. x = 4

  26. 2x = (4x + 8) 1 1 2 2 (2x)2 = [(4x + 8) ]2 Example 4B: Solving Equations with Rational Exponents Step 1 Solve for x. Raise both sides to the reciprocal power. Simplify. 4x2 = 4x + 8 Write in standard form. 4x2 – 4x – 8 = 0 4(x2 – x – 2) = 0 Factor out the GCF, 4. 4(x – 2)(x + 1) = 0 Factor. 4 ≠ 0, x –2 = 0 or x + 1 = 0 Solve for x. x = 2 or x = –1

  27. 1 1 1 1 1 1 2 2 2 2 2 2 2x = (4x + 8) 2x = (4x + 8) 2(–1) (4(–1) + 8) 4 4 –2 2 x  –2 4 4 16 2(2) (4(2) + 8) Example 4B Continued Step 2 Use substitution to check for extraneous solutions. The only solution is x = 2.

  28. 1 3 Check It Out! Example 4a Solve each equation. (x + 5) = 3 Write in radical form. Cube both sides. x + 5 = 27 Simplify. Solve for x. x = 22

  29. 1 1 2 2 [(2x + 15) ]2 =(x)2 Check It Out! Example 4b (2x + 15) = x Step 1Solve for x. Raise both sides to the reciprocal power. Simplify. 2x + 15 = x2 Write in standard form. x2 – 2x – 15 = 0 (x – 5)(x + 3) = 0 Factor. x –5 = 0 or x + 3 = 0 Solve for x. x = 5 or x = –3

  30. 1 1 1 1 1 1 2 2 2 2 2 2 (2x + 15) = x (2x + 15) = x (2(5) + 15) 5 (2(–3) + 15) –3 3 –3 x 5 5  (10 + 15) 5 (–6 + 15) –3 Check It Out! Example 4b Continued Step 2 Use substitution to check for extraneous solutions. The only solution is x = 5.

  31. 1 1 2 2 [3(x + 6) ]2 = (9)2 Check It Out! Example 4c 3(x + 6) = 9 Raise both sides to the reciprocal power. 9(x + 6) = 81 Simplify. Distribute 9. 9x + 54 = 81 9x = 27 Simplify. Solve for x. x = 3

  32. Check It Out! Example 6 The speed s in miles per hour that a car is traveling when it goes into a skid can be estimated by using the formula s = , where f is the coefficient of friction and d is the length of the skid marks in feet. 30 fd A car skids to a stop on a street with a speed limit of 30 mi/h. The skid marks measure 35 ft, and the coefficient of friction was 0.7. Was the car speeding? Explain. Use the formula to determine the greatest possible length of the driver’s skid marks if he was traveling 30 mi/h.

  33. Check It Out! Example 6 Continued Substitute 30 for s and 0.7 for f. Simplify. Square both sides. 900 = 21d Simplify. 43 ≈ d Solve for d. If the car were traveling 30 mi/h, its skid marks would have measured about 43 ft. Because the actual skid marks measure less than 43 ft, the car was not speeding.

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