How to find x in #sqrt(x+3)+sqrt(2-x)=3#? Please show complete solution. Thanks so much.

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Answer 1 · 2017-10-10T05:49:48

#x=1,\qquad x=-2#

#\sqrt{x+3}+\sqrt{2-x}=3#

#\Rightarrow\sqrt{x+3}=3-\sqrt{2x-2}#

#\Rightarrow(\sqrt{x+3})^2=(2-\sqrt{2-x})^2#

#\Rightarrowx+3=11-6\sqrt{2x}-x#

#\Rightarrow2x-8=-6\sqrt{2-x}#

#\Rightarrow(2x-8)^2=(-6\sqrt{2-x})^2#

#\Rightarrow 4x-32x+64=72-36x#

#\Rightarrow4x^2+4x-9=0#

#x=\frac{-4+\sqrt{4^2-4(4)(-8)}}{2\cdot4}#

#=1#

Quadratic Formula (Minus):

#x=\frac{-4-\sqrt{4^2-4(4)(-8)}}{2\cdot 4}#

#=-2#

#\therefore\quad x=1,\qquad x=-2#