Question

square root>(3x + 1)− square root>(7x + 8)= 1

Answer 1

√(3x+1) - √(7x+8) = 1 , clearly x≥-1/3
√(3x+1) = 1 + √(7x+8)
square both sides

3x+1 = 1 + 2√(7x+8) + 7x+8
-4x -8 = 2√(7x+8)
-2x-4 = √7x+8)
square both sides again
4x^2 + 16x + 16 = 7x+8
4x^2 + 9x + 8 = 0
x = (-9 ± √-47)/8

no real solution.

Answer 2

I'm sorry I wrote the equation wrong 3ã(x+1)-ã(7x+8)=1

Answer 3

square roots to replace the a's

Answer 4

so
3√(x+1) - √(7x+8) = 1 , clearly x≥-1
3√(x+1) = 1 + √(7x+8)
square both sides

9(x+1) = 1 + 2√(7x+8) + 7x+8
9x+9 = 1 + 2√(7x+8) + 7x+8
2x = 2√(7x+8)
square again
4x^2= 4(7x+8)
4x^2 - 28x - 32 = 0
x^2 - 7x - 8 = 0
(x-8)(x+1) = 0
x=8 or x=-1

since we squared, all answers must be checked

if x = 8
LS = 3√9 - √(7(8) + 8)
= 9 - 8
= RS

if x = -1
LS = 3√0 - √(-7)+8))
= 0-1
≠ RS

So x = 8