In the equation, $x - \sqrt{x - 4} = 4$ $x-sqrt(x-4)=4$ , how do we solve for x, showing the complete solution? Thanks

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Answer 1 · 2017-10-10T06:46:21

x = 4 0r 5

Rewrite the equation as, $- \sqrt{x - 4} = 4 - x$ $- sqrt(x-4) = 4 - x$

squaring both sides, we get
${[- \sqrt{x - 4}]}^{2} = {(4 - x)}^{2}$ $[-sqrt(x-4)]^2 = (4-x)^2$

$\Rightarrow x - 4 = 16 - 8 x + x^{2}$ $rArr x - 4 = 16 - 8x + x^2$

$\Rightarrow x^{2} - 8 x - x + 16 + 4 = 0$ $rArr x^2-8x-x+16+4 =0$

$\Rightarrow x^{2} - 9 x + 20 = 0$ $rArr x^2-9x+20 = 0$

$\Rightarrow x^{2} - 4 x - 5 x + 20 = 0$ $rArr x^2-4x-5x+20 =0$

$\Rightarrow x (x - 4) - 5 (x - 4) = 0$ $rArr x(x-4)-5(x-4)=0$

$\Rightarrow (x - 4) (x - 5) = 0$ $rArr(x-4)(x-5)=0$

$\Rightarrow (x - 4) = 0 and (x - 5) = 0$ $rArr (x-4) = 0 and (x-5)=0$

$\Rightarrow x = 4 or 5$ $rArr x = 4 or 5$

In the equation, x−√x−4=4x-sqrt(x-4)=4, how do we solve for x, showing the complete solution? Thanks