ALGEBRA 1 Solving Radical Equations 1)Isolate the radical 2)Square both sides 3)Simplify.

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Presentation transcript:

ALGEBRA 1 Solving Radical Equations 1)Isolate the radical 2)Square both sides 3)Simplify

ALGEBRA 1 Solve each equation. Check your answers. a. x – 5 = 4 x = 9Isolate the radical on the left side of the equation. ( x) 2 = 9 2 Square each side. x = 81 x – 5 = 4 4 = 4 Check: x – 5 = 4 81 – 5 4 Substitute 81 for x. 9 – 5 4 Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 b. x – 5 = 4 x – 5 = 16Solve for x. x = 21 ( x – 5) 2 = 4 2 Square each side. (continued) Check: x – 5 = 4 4 = 4 16 = 4 21– 5 = 4Substitute 21 for x. Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 Solving Radical Equations LESSON 10-4 Additional Examples On a roller coaster ride, your speed in a loop depends on the height of the hill you have just come down and the radius of the loop in feet. The equation v = 8 h – 2r gives the velocity v in feet per second of a car at the top of the loop.

ALGEBRA 1 The loop on a roller coaster ride has a radius of 18 ft. Your car has a velocity of 120 ft/s at the top of the loop. How high is the hill of the loop you have just come down before going into the loop? Solve v = 8 h – 2r for h when v = 120 and r = = 8 h – 2(18) Substitute 120 for v and 18 for r. = Divide each side by 8 to isolate the radical. 8 h – 2(18) (15) 2 = ( h – 36) 2 Square both sides. 225 = h – = h The hill is 261 ft high. (continued) 15 = h – 36Simplify. Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 Roller Coasters Banshee at Kings Island Batman: The Ride

ALGEBRA 1 Solve 3x – 4 = 2x + 3. ( 3x – 4) 2 = ( 2x + 3) 2 Square both sides. 3x – 4 = 2x + 3Simplify. 3x = 2x + 7Add 4 to each side. x = 7Subtract 2x from each side. The solution is 7. Check: 3x – 4 = 2x = 17 3(7) – 4 2(7) + 3Substitute 7 for x. Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 (x) 2 = ( x + 12) 2 Square both sides. x 2 = x + 12 x 2 – x – 12 = 0 Simplify. The solution to the original equation is 4. The value –3 is an extraneous solution. Solve x = x (x – 4)(x + 3) = 0 Solve the quadratic equation by factoring. (x – 4) = 0 or (x + 3)= 0 Use the Zero–Product Property. x = 4 or x = –3 Solve for x. Check: x = x –3 – = 4 –3 ≠ 3 Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 Solve 3x + 8 = 2. 3x = –6 Subtract 8 from each side. ( 3x) 2 = (–6) 2 Square both sides. 3x = 36 x = 12 3x + 8 = 2 has no solution. Check: 3x + 8 = 2 3(12) Substitute 12 for x ≠ 2 x = 12 does not solve the original equation. Solving Radical Equations LESSON 10-4 Additional Examples

ALGEBRA 1 Practice Go to your assigned groups and complete the following problems in the textbook: p. 636 #7 - 24, 37, 41, 49