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):

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

$=-2$

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

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