CONFIDENTIAL 1 Algebra1 Solving Radical Equations

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 WORDSNUMBERSALGEBRA You can square both sides of an equation, and the resulting equation is still true. 3 = (3)2 + (1 + 2) 2 9 = 9 If a and b are real numbers and a = b, then a 2 = b 2.

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

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

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 = 7 2 x = 49 Subtract 3 from both sides. Square both sides. Check: Substitute 49 for x in the original equation. Simplify. √x + 3 = 10 √(49)

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

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

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. A)3√x = 21 Method 1: √x = 7 √(x) 2 = 7 2 x = 49 Method 2: 3√x = 21 (3√(x)) 2 = (21) 2 9x = 441 x = 49 Divide both sides by 3. Square both sides. Check: Substitute 49 for x in the original equation. Simplify. 3√x = 21 3√(49) 21 3(7) Divide both sides by 9. Square both sides.

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. Check: Substitute 225 for x in the original equation. Simplify. √x = 5 3 √(225) Multiply both sides by 9. Square both sides.

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. A)√(x + 1) = √3 (√(x + 1)) 2 = (√3) 2 x + 1 = 3 x = 2 Square both sides. Check: Substitute 2 for x in the original equation. Simplify. √(x + 1) = √3 √(2 + 1) √3 √3 √3 Subtract 1 from both sides.

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

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. 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. x = 3 x 2 = 9 √(x) 2 = √9 x = 3 or x = -3

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 = x 2 x 2 + 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 6) A rectangle has an area of 15 cm 2. 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? Now you try!

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 cm 2. 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 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. Let’s review

CONFIDENTIAL 32 Power Property of Equality WORDSNUMBERSALGEBRA You can square both sides of an equation, and the resulting equation is still true. 3 = (3)2 + (1 + 2) 2 9 = 9 If a and b are real numbers and a = b, then a 2 = b 2.

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

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

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

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. 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. x = 3 x 2 = 9 √(x) 2 = √9 x = 3 or x = -3

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 = x 2 x 2 + 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!