How do you solve $(-3-4x)^(1/2)-(-2-2x)^(1/2)=1$ ?

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Answer 1 · 2018-02-03T23:52:23

$x=3+-sqrt(29)/2$

$sqrt(-3-4x)-sqrt(-2-2x)=1$

Square both sides,

$-3-4x-2sqrt((-3-4x)(-2-2x))-2-2x=1$

Multiply And Combine like terms.

$-6x-5-2sqrt(8x^2+14x+6)=1$

Add $[6x+5]$

$-2sqrt(8x^2+14x+6)=6x+6$

Square both Sides again And Multiply.

$32x^2+56x+24=36x^2+72x+36$

Subtract $[36x^2+72x+36]$

$-4x^2-24x-28=0$

Use quadratic Formula And simplify again.

$x=(24+-sqrt(576-112))/-8=3+-sqrt(29)/2$

How do you solve (-3-4x)^(1/2)-(-2-2x)^(1/2)=1(-3-4x)^(1/2)-(-2-2x)^(1/2)=1?