Question

If $\sqrt{3-2x}+\sqrt{7+2x}=4$, then find the possible value of $x$?

• A. $1,-3$
• B. $2,3$
• C. $2,-3$
• D. $-1,3$

Answer 1

The correct option is A $1,-3$

Given, $\sqrt{3-2x}+\sqrt{7+2x}=4$

or, $\sqrt{3-2x}=4-\sqrt{7+2x}$

Squaring on both sides, we get

$(\sqrt{3-2x}{)}^2=(4-\sqrt{7+2x}{)}^2$

$\Rightarrow 3-2x=16+7+2x-8\sqrt{7+2x}$

$\Rightarrow 4x+20=8\sqrt{7+2x}$

$\Rightarrow x+5=2\sqrt{7+2x}$

Again squaring on both sides, we get

$(x+5{)}^2=(2\sqrt{7+2x}{)}^2$

$\Rightarrow x^2+10x+25=4(7+2x)$

$\Rightarrow x^2+10x+25=28+8x$

$\Rightarrow x^2+2x-3=0$

$\Rightarrow x^2+3x-x-3=0$

$\Rightarrow (x+3)(x-1)=0$

$\Rightarrow x=-3$ or $x=1$

Possible value of $x=1,-3$

Hence, A is the correct option.