help algabra II Solve each equation Check for extraneous sol

help algabra II

Solve each equation. Check for extraneous solutions 2 + squareroot x + 5 = 4 Electricity the powei P, in watts, that a circular

Solution

47. 4 ( 3x -3)^(2/3) = 36

( 3x -3)^(2/3) = 36/4 =9

(3x -3)^(2/3) = 9

take cube of both sides: (3x -3)^2 = 729

taking square root of both sides: (3x -3) = 27

3x = 3+ 27

x = 10

Check x= 10 is extraneous solution or not :

( 3x -3)^(2/3) = 9

( 3*10 -3)^(2/3) =9

(27)^2/3 =9

3^2 =9

Hence x =10 is a solution of the equation

49. sqrt( x +6) +2 = x+6

sqrt(x+6) = x+4

taking square of both sides:

(x+6) = (x+4)^2

x+6 = x^2 +16 +8x

x^2 +7x +10 =0

Using quadratic root formula for solution:

x = ( -7 + /- sqrt( 49- 40)/2

= ( -7 + / - 3)/2

x = -5 , -2

Checking whether x = -5 and x= -2 is extraneous solution:

sqrt( x +6) +2 = x+6

we can check that sqrt( -5 +6) +2 = -5 +6

sqrt(1) +2 = 1. So, x=-5 is an extraneous solution

x = -2 ; sqrt( -2 +6) +2 = -2 +6

2 +2 = 4

x = -2 is a solution of the equation.

51.

sqrt(2x +9) - sqrtx = 3

sqrt(2x +9) = 3 +sqrtx

taking square root of both sides:

(2x +9) = 9 +x +6xqrtx

x = 6sqrt(x)

squaring both sides: x^2 = 36x

x^2 -36x =0

x( x-36) =0

x =0 and x=36 are solutions

Check whether they are extraneous solution:

x = 0 : sqrt(0 +9) +sqrt0 = 3

3 =3 So, x= 0 is a solution

x = 36 ; sqrt( 72 +9) -sqrt36 = 3

9 -6 = 3

So, x= 36 is asolution.

hence x =0, 36 are solutions to the equation