Solve a radical equation
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- Stuart Hampton
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1 EXAMPLE 1 Solve a radical equation Solve 3 = 3. x+7 3 x+7 = 3 Write original equation. 3 ( x+7 ) 3 = 3 3 x+7 = 7 x = 0 x = 10 Cube each side to eliminate the radical. Subtract 7 from each side. Divide each side by.
2 EXAMPLE 1 CHECK Solve a radical equation Check x = 10 in the original equation. 3 (10) = = 3 3 Substitute 10 for x. 3 = 3 Solution checks.
3 GUIDED PRACTICE for Example 1 Solve equation. Check your solution x 9 = 1 3 x 9 = 1 Write original equation. 3 x = 8 Add 9 to each side. ( 3 x ) 3 = (8) 3 Use each side to eliminate the radical. x = 51
4 GUIDED PRACTICE for Example 1 Solve equation. Check your solution.. ( x+5 ) = 4 ( x+5 ) = 4 Write original equation. ( ) = 4 x+5 x + 5 = 16 Square each side to eliminate the radical. x = 9 Subtract 5 from each side.
5 GUIDED PRACTICE for Example 1 Solve equation. Check your solution. 3. ( 3 x 3 ) = 4 ( 3 x 3 ) = 4 Write original equation. 3 x 3 = Divided from each side. ( 3 x 3 ) 3 = 3 Cube each side to eliminate the radical. x 3 = 8 x = 11 Add 3 to each side.
6 EXAMPLE Solve a radical equation given a function Wind Velocity In a hurricane, the mean sustained wind velocity v (in meters per second) is given by v(p) = p where p is 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.
7 EXAMPLE Solve a radical equation given a function SOLUTION v(p) = p Write given function = p Substitute 54.5 for v(p) p Divide each side by 6.3. (8.65) 1013 p Square each side p 938. p Subtract 1013 from each side p Divide each side by 1.
8 EXAMPLE Solve a radical equation given a function ANSWER The air pressure at the center of the hurricane is about 938 millibars.
9 GUIDED PRACTICE for Example 4. What If Use the function in Example to estimate the air pressure at the center of a hurricane when the mean sustained wind velocity is 48.3 meters per second. SOLUTION v(p) = p Write given function = p Substitute 48.3 for v(p) p Divide each side by 6.3. (7.67) 1013 p Square each side.
10 GUIDED PRACTICE for Example p 954 p Subtract 1013 from each side. 954 p Divide each side by 1. ANSWER The air pressure at the center of the hurricane is about 954 mille bars.
11 EXAMPLE 3 Standardized Test Practice SOLUTION 4x /3 = 36 Write original equation. x /3 = 9 Divide each side by 4. (x /3 ) 3/ = 9 3/ x = ±7 ANSWER The correct answer is D. 3 Raise each side to the power.
12 EXAMPLE 4 Solve an equation with a rational exponent Solve (x + ) 3/4 1 = 7. (x + ) 3/4 1 = 7 (x + ) 3/4 = 8 (x + ) 3/4 4/3 = 8 4/3 x + = (8 1/3 ) 4 x + = 4 x + = 16 x = 14 Write original equation. Add 1 to each side. Raise each side to the power 4. 3 Apply properties of exponents. Subtract from each side.
13 EXAMPLE 4 Solve an equation with a rational exponent ANSWER The solution is 14. Check this in the original equation.
14 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution. 5. 3x 3/ = 375 3x 3/ = 375 Write original equation. x 3/ = 15 Divide each side by 3. (x 3/ ) /3 = (15) /3 Raise each side to the power. 3 x = 5
15 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution. 6. x 3/4 = 16 x 3/4 = 16 Write original equation. x 3/4 = 8 Divide each side by. (x 3/4 ) 4/3 = 8 4 4/3 Raise each side to the power. 3 x = (8 1/3 ) 4 Apply properties of exponent. x = 16
16 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution x1/5 = 3 x1/5 = x 1/5 = 3 Write original equation. Divide each side by /3. (x 1/5 ) 5 = 3 5 Raise each side to the power 5. x = 43
17 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution. 8. (x + 3) 5/ = 3 (x + 3) 5/ = 3 Write original equation. [(x+ 3) 5/ ] /5 = 3 /5 Raise each side to the power /5. x + 3 = (3 1 ) 5 x + 3 = 4 Apply properties of exponent. x = 1
18 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution. 9. (x 5) 4/3 = 81 (x 5) 4/3 = 81 Write original equation. [(x 5) 4/3 ] 3/4 = (81) 3/4 Raise each side to the power 3/4. x 5 = (81 1/4 ) 3 x 5 = ±3 3 x 5 = ±7 x 5 = 7 or x 5 = 7 x = 3 or x = Apply properties of exponent. Let (x 5) equal 7 and 7. Subtract 5 from both sides of each equation.
19 GUIDED PRACTICE for Examples 3 and 4 Solve the equation. Check your solution. 10. (x + ) /3 + 3 = 7 (x + ) /3 +3 = 7 (x + ) /3 = 4 [(x + ) /3 ] 3/ = 4 3/ Write original equation. Subtract each side by 3. Raise each side to the power 3/. x + = (4 1/ ) 3 x + = 8 or x + = 8 x = 10 or 6 Apply properties of exponent.
20 EXAMPLE 5 Solve an equation with an extraneous solution Solve x + 1 = 7x x + 1 = 7x + 15 (x + 1) = ( 7x + 15) x + x + 1 = 7x + 15 x 5x 14 = 0 (x 7)(x + ) = 0 x 7 = 0 or x + = 0 Write original equation. Square each side. Expand left side and simplify right side. Write in standard form. Factor. Zero-product property x = 7 or x = Solve for x.
21 EXAMPLE 5 Solve an equation with an extraneous solution CHECK 1 Check x = 7 in the original equation. x + 1 = 7x = 7(7) = 64 Check x = in the original equation. x + 1 = 7x = 7( ) = 1 8 = 8 1 = 1 / ANSWER The only solution is 7. (The apparent solution is extraneous.)
22 EXAMPLE 6 Solve an equation with two radicals Solve x = 3 x. SOLUTION METHOD 1 Solve using algebra. x = 3 x x = 3 x x + + x + +1 = 3 x x + = x Write original equation. Square each side. Expand left side and simplify right side. Isolate radical expression.
23 EXAMPLE 6 Solve an equation with two radicals x + = x x + = ( x) Divide each side by. Square each side again. x + = x 0 = x x Write in standard form. 0 = (x )(x + 1) Factor. x = 0 x + 1 = 0 or x = or x = 1 Zero-product property. Solve for x.
24 EXAMPLE 6 Solve an equation with two radicals Check x = in the original equation. Check x = 1 in the original equation. x + +1 = 3 x x + +1 = 3 x + +1 = = 3 ( 1) 4 +1 = = 4 3 = 1 / = ANSWER The only solution is 1. (The apparent solution is extraneous.)
25 EXAMPLE 6 Solve an equation with two radicals METHOD Use: a graph to solve the equation. Use a graphing calculator to graph y 1 = x and y = 3 x. Then find the intersection points of the two graphs by using the intersect feature. You will find that the only point of intersection is (1, ). Therefore, 1 is the only solution of the equation x = 3 x
26 GUIDED PRACTICE for Examples 5 and 6 Solve the equation. Check for extraneous solutions 11. x 1 = 1 x 4 x 1 = 1 x 4 (x + 1 ) 1 = x 4 x x + 1 = 1 x 4 4 x 5 x + 1 = x 5x + 1 = 0 ( ) Write original equation. Square each side. Expand left side and simplify right side. Write in standard form. (4x 1)(x 1) = 0 Factor.
27 GUIDED PRACTICE for Examples 5 and 6 4x 1 = 0 or x 1 = 0 x = 1 or x = 1 4 Check x = 1 in the original equation. x 1 1 = 1 x = x 1 1 = 1 = Zero-product property. Solve for x. 1 Check x = in the 4 original equation. x = 1 = 1 = 4 1 = x 4 1 x The only solution is 1 (the apparent solution 1/4 is extraneous) 1 16
28 GUIDED PRACTICE for Examples 5 and 6 Solve the equation. Check for extraneous solutions 1. 10x + 9 = x x + 9 = x + 3 Write original equation. ( ) 10x + 9 = (x + 3) 10x + 9 = x + 6x +9 x 4x = 0 x (x 4) = 0 Square each side. Expand right side and simplify left side. Write in standard form. Factor. (x 4) = 0 x = 4 or or x = 0 x = 0 Zero-product property. Solve for x.
29 GUIDED PRACTICE for Examples 5 and 6 Check x = 4 in the original equation. 10x + 9 = x x 4+ 9 = = 7 49 = 7 7 = 7 Check x = 0 in the original equation. 10x + 9 = x x 0+ 9 = = 3 3 = 3 The solution are 4 and 0.
30 GUIDED PRACTICE for Examples 5 and 6 Solve the equation. Check for extraneous solutions 13. x + 5 x + 7 = x + 5 = x + 7 ( x + 5 ) ( ) = x + 7 Write original equation. Square each side. x + 5 = x + 7 x = 0 x = Simplify both the sides.
31 GUIDED PRACTICE for Examples 5 and 6 Check x = in the original equation x + 5 x + 7 =. + 5 = = 9 3 = 3 The solution is.
32 GUIDED PRACTICE for Examples 5 and 6 Solve the equation. Check for extraneous solutions 14. x + 6 = x Solve x + 6 x = Write original equation. ( x + 6 ) = ( x ) x x = x Square each side. Simplify each side. 4 x + 6 = 1 x + 6 = 3
33 GUIDED PRACTICE for Examples 5 and 6 x + 6 = 3 Divide each side by 4. ( ) x + 6 = 3 Square each side. x + 6 = 9 x = 3
34 GUIDED PRACTICE for Examples 5 and 6 Check x = 3 in the original equation x + 6 = x = 3 9 = 1 1 = 1
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