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#