Question

Someone please help me :((

Answer 1

8x^2-3 = sqrt(16x+9)
[8x^2-3]^2 = [sqrt(16x+9)]^2 ... square both sides
64x^4-48x^2+9 = 16x+9
64x^4-48x^2+9-16x-9 = 0
64x^4-48x^2-16x = 0
16x(4x^3-3x-1) = 0
16x(x-1)(2x+1)^2 = 0
16x=0 or x-1=0 or (2x+1)^2 = 0
x=0 or x=1 or x = -1/2

The possible solutions are x=0 or x=1 or x = -1/2

We need to check all the possible solutions

Checking x=0
8x^2-3 = sqrt(16x+9)
8(0)^2-3 = sqrt(16*0+9)
-3 = 3
The equation is false so x=0 is extraneous (not a real solution)

Checking x=1
8x^2-3 = sqrt(16x+9)
8(1)^2-3 = sqrt(16*1+9)
5 = 5
Equation is true. The value x=1 is a solution

Checking x=-1/2
8x^2-3 = sqrt(16x+9)
8(-1/2)^2-3 = sqrt(16(-1/2)+9)
-1 = 1
The equation is false so x=-1/2 is extraneous (not a real solution)

Therefore, the only answer is choice A) 1