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Algebra1 Solving Radical Equations

Algebra1 Solving Radical Equations. Warm Up. 1) A dessert menu offers 6 different selections. The restaurant offers a dessert sampler that includes small portions of any 4 different choices from the dessert menu. How many different dessert samplers are possible?. Solving Radical Equations.

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Algebra1 Solving Radical Equations

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  1. Algebra1Solving RadicalEquations CONFIDENTIAL

  2. Warm Up 1) A dessert menu offers 6 different selections. The restaurant offers a dessert sampler that includes small portions of any 4 different choices from the dessert menu. How many different dessert samplers are possible? CONFIDENTIAL

  3. Solving Radical Equations A radical equation is an equation that contains a variable within a radical. In this course, you will only study radical equations that contain square roots. Recall that you use inverse operations to solve equations. For nonnegative numbers, squaring and taking the square root are inverse operations. When an equation contains a variable within a square root, square both sides of the equation to solve. CONFIDENTIAL

  4. Power Property of Equality CONFIDENTIAL

  5. Solving Simple Radical Equations Solve each equation. Check your answer. A) √x = 8 √(x)2 = 82 x = 64 Square both sides. Check: √x = 8 √(64) 8 8 8 Substitute 64 for x in the original equation. Simplify. CONFIDENTIAL

  6. B) 6 = √(4x) √62 = √(4x)2 36 = x 9 = x Square both sides. Divide both sides by 4. Check: 6 = √(4x) 6 √(4(9)) 6 √(36) 6 6 Substitute 9 for x in the original equation. Simplify. CONFIDENTIAL

  7. Now you try! Multiply. Write each product in simplest form. 1a) √x = 6 1b) 9 = √(27x) 1c) √(3x) = 1 CONFIDENTIAL

  8. Some square-root equations do not have the square root isolated. To solve these equations, you may have to isolate the square root before squaring both sides. You can do this by using one or more inverse operations. CONFIDENTIAL

  9. Solving Radical Equations by Adding or Subtracting Solve each equation. Check your answer. A) √x + 3 = 10 √x = 7 √(x)2 = 72 x = 49 Subtract 3 from both sides. Square both sides. Check: √x + 3 = 10 √(49) + 3 10 7 + 3 10 10 10 Substitute 49 for x in the original equation. Simplify. CONFIDENTIAL

  10. B) √(x – 5) = 4 √(x – 5)2 = 42 (x – 5) =16 x = 21 Square both sides. Add 5 to both sides. Check: √(x – 5) = 4 √(21 – 5) 4 √(16) 4 4 4 Substitute 21 for x in the original equation. Simplify. CONFIDENTIAL

  11. C) √(2x – 1) + 4 = 7 √(2x – 1) = 3 (√(2x – 1))2 = (3)2 2x - 1 = 9 2x = 10 x = 5 Subtract 4 from both sides. Square both sides. Add 1 to both sides. Divide both sides by 2. √(2x – 1) + 4 = 7 √(2(5) – 1) + 4 7 √(10 - 1) + 4 7 √9 + 4 7 3 + 4 7 7 7 Substitute 5 for x in the original equation. Check: Simplify. CONFIDENTIAL

  12. Now you try! Solve each equation. Check your answer. 2a) √x - 2 = 1 2b) √(x + 7) = 5 2c) √(3x + 7) - 1 = 3 CONFIDENTIAL

  13. Solving Radical Equations by Multiplying or Dividing Solve each equation. Check your answer. • 3√x = 21 • Method 1: • √x = 7 • √(x)2 = 72 • x = 49 • Method 2: • 3√x = 21 • (3√(x))2 = (21)2 • 9x = 441 • x = 49 Divide both sides by 3. Square both sides. Square both sides. Divide both sides by 9. 3√x = 21 3√(49) 21 3(7) 21 21 21 Substitute 49 for x in the original equation. Check: Simplify. CONFIDENTIAL

  14. B) √x = 5 3 Method 1: √x = 15 √(x)2 = (15)2 x = 225 Method 2: √x = (5)2 3 x = 25 9 x = 225 Multiply both sides by 3. Square both sides. Square both sides. Multiply both sides by 9. √x = 5 3 √(225) 5 3 15 5 3 5 5 Check: Substitute 225 for x in the original equation. Simplify. CONFIDENTIAL

  15. Now you try! Solve each equation. Check your answer. 3a) 2√x = 22 3b) 2 = √x 4 3c) 2√x = 4 5 CONFIDENTIAL

  16. Solving Radical Equations with Square Roots on Both Sides Solve each equation. Check your answer. • √(x + 1) = √3 • (√(x + 1))2 = (√3)2 • x + 1 = 3 • x = 2 Square both sides. Subtract 1 from both sides. √(x + 1) = √3 √(2 + 1) √3 √3 √3 Substitute 2 for x in the original equation. Check: Simplify. CONFIDENTIAL

  17. B) √(x + 8) - √(3x) = 0 √(x + 8) = √(3x) (√(x + 8))2 = (√(3x))2 x + 8 = 3x 2x = 8 x = 4 Add √(3x) from both sides. Square both sides. Subtract x from both sides. Divide both sides by 2. Check: CONFIDENTIAL

  18. Now you try! Solve each equation. Check your answer. CONFIDENTIAL

  19. Squaring both sides of an equation may result in an extraneous solution — a number that is not a solution of the original equation. x = 3 x2 = 9 √(x)2 = √9 x = 3 or x = -3 Suppose your original equation is x = 3. Square both sides. Now you have a new equation. Solve this new equation for x by taking the square root of both sides. Now there are two solutions. One (x = 3) is the original equation. The other (x = -3) is extraneous—it is not a solution of the original equation. Because of extraneous solutions, it is important to check your answers. CONFIDENTIAL

  20. Solving Radical Equations with Square Roots on Both Sides Solve √(6 – x) = x. Check your answer. (√(6 – x))2 = (x)2 6 – x = x2 x2 + x - 6 = 0 (x - 2) (x + 3) = 0 x - 2 = 0 or x + 3 = 0 x = 2 or x = -3 Square both sides. Write in standard form. Factor. Zero-Product Property Solve for x. CONFIDENTIAL

  21. Check: Substitute 2 for x in the equation. Substitute -3 for x in the equation. -3 does not check; it is extraneous. The only solution is 2. CONFIDENTIAL

  22. Now you try! Multiply. Write each product in simplest form. CONFIDENTIAL

  23. Geometry Application A rectangle has an area of 52 square feet. Its length is 13 feet, and its width is √x feet. What is the value of x? What is the width of the rectangle? Use the formula for area of a rectangle. Substitute 52 for A, 13 for l, and √ x for w. Divide both sides by 13. Square both sides. CONFIDENTIAL

  24. Check: Substitute 16 for x in the equation. The value of x is 16. The width of the rectangle is √(16) = 4 feet. CONFIDENTIAL

  25. Now you try! 6) A rectangle has an area of 15 cm2 . Its width is 5 cm, and its length is √(x + 1) cm. What is the value of x? What is the length of the rectangle? CONFIDENTIAL

  26. Assessment 1) Is x = √3 a radical equation? Why or why not? CONFIDENTIAL

  27. Solve each equation. Check your answer. CONFIDENTIAL

  28. Solve each equation. Check your answer. 6) 7) CONFIDENTIAL

  29. Solve each equation. Check your answer. 8) 9) CONFIDENTIAL

  30. 10) A trapezoid has an area of 14 cm2 . The length of one base is 4 cm and the length of the other base is 10 cm. The height is √(2x + 3) cm. What is the value of x? What is the height of the trapezoid? CONFIDENTIAL

  31. Let’s review Solving Radical Equations A radical equation is an equation that contains a variable within a radical. In this course, you will only study radical equations that contain square roots. Recall that you use inverse operations to solve equations. For nonnegative numbers, squaring and taking the square root are inverse operations. When an equation contains a variable within a square root, square both sides of the equation to solve. CONFIDENTIAL

  32. Power Property of Equality CONFIDENTIAL

  33. Solving Radical Equations by Adding or Subtracting Solve each equation. Check your answer. A) √x + 3 = 10 √x = 7 √(x)2 = 72 x = 49 Subtract 3 from both sides. Square both sides. Check: √x + 3 = 10 √(49) + 3 10 7 + 3 10 10 10 Substitute 49 for x in the original equation. Simplify. CONFIDENTIAL

  34. Solving Radical Equations by Multiplying or Dividing Solve each equation. Check your answer. • 3√x = 21 • Method 1: • √x = 7 • √(x)2 = 72 • x = 49 • Method 2: • 3√x = 21 • (3√(x))2 = (21)2 • 9x = 441 • x = 49 Divide both sides by 3. Square both sides. Square both sides. Divide both sides by 9. 3√x = 21 3√(49) 21 3(7) 21 21 21 Substitute 49 for x in the original equation. Check: Simplify. CONFIDENTIAL

  35. Solving Radical Equations with Square Roots on Both Sides Solve each equation. Check your answer. • √(x + 1) = √3 • (√(x + 1))2 = (√3)2 • x + 1 = 3 • x = 2 Square both sides. Subtract 1 from both sides. √(x + 1) = √3 √(2 + 1) √3 √3 √3 Substitute 2 for x in the original equation. Check: Simplify. CONFIDENTIAL

  36. Squaring both sides of an equation may result in an extraneous solution — a number that is not a solution of the original equation. x = 3 x2 = 9 √(x)2 = √9 x = 3 or x = -3 Suppose your original equation is x = 3. Square both sides. Now you have a new equation. Solve this new equation for x by taking the square root of both sides. Now there are two solutions. One (x = 3) is the original equation. The other (x = -3) is extraneous—it is not a solution of the original equation. Because of extraneous solutions, it is important to check your answers. CONFIDENTIAL

  37. Solving Radical Equations with Square Roots on Both Sides Solve √(6 – x) = x. Check your answer. (√(6 – x))2 = (x)2 6 – x = x2 x2 + x - 6 = 0 (x - 2) (x + 3) = 0 x - 2 = 0 or x + 3 = 0 x = 2 or x = -3 Square both sides. Write in standard form. Factor. Zero-Product Property Solve for x. CONFIDENTIAL

  38. Geometry Application A rectangle has an area of 52 square feet. Its length is 13 feet, and its width is √x feet. What is the value of x? What is the width of the rectangle? Use the formula for area of a rectangle. Substitute 52 for A, 13 for l, and √ x for w. Divide both sides by 13. Square both sides. CONFIDENTIAL

  39. You did a great job today! CONFIDENTIAL

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