How do you solve $\sqrt{x - 7} = 7 - \sqrt{x}$ $sqrt(x-7) = 7 - sqrt(x)$ ?

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Answer 1 · 2016-04-03T04:56:33

$x = 16$ $x=16$

$\sqrt{x - 7} = 7 - \sqrt{x}$ $color(blue)(sqrt(x-7)=7-sqrtx$

Take the square of both sides to remove the radical sign in the l.h.s

$\to {(\sqrt{x - 7})}^{2} = {(7 - \sqrt{x})}^{2}$ $rarr(sqrt(x-7))^2=(7-sqrtx)^2$

$\to x - 7 = {(7 - \sqrt{x})}^{2}$ $rarrx-7=(7-sqrtx)^2$

Use the formula ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$ $color(brown)((a-b)^2=a^2-2ab+b^2$

$\to x - 7 = 7^{2} - 2 (7) (\sqrt{x}) + {\sqrt{x}}^{2}$ $rarrx-7=7^2-2(7)(sqrtx)+sqrtx^2$

$\to x - 7 = 49 - 14 \sqrt{x} + x$ $rarrx-7=49-14sqrtx+x$

Subtract $x$ $x$ both sides

$rarrcancelx-7cancel(-x)=49-14sqrtx+cancel(x-x$

$rarr-7=49-14sqrtx$

Subtract $49$ both sides

$rarr-7-49=cancel49-14sqrtxcancel(-49$

$rarr-56=-14sqrtx$

Divide both sides by $-14$

$rarr(-56)/-14=(cancel(-14)sqrtx)/cancel(-14$

$rarr4=sqrtx$

Square both sides

$rarr4^2=sqrtx^2$

$color(green)(rarr16=x$

How do you solve sqrt(x-7) = 7 - sqrt(x)√x−7=7−√xsqrt(x-7) = 7 - sqrt(x)?