The sum of the solutions of the equation |sqr | vedclass

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(C) Let $t = \sqrt{x}$,where $t > 0$.The equation becomes $|t-2| + t(t-4) + 2 = 0$.$|t-2| + t^2 - 4t + 2 = 0$.We can rewrite $t^2 - 4t + 2$ as $(t^2 -

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Let $t = \sqrt{x}$,where $t > 0$.The equation becomes $|t-2| + t(t-4) + 2 = 0$.$|t-2| + t^2 - 4t + 2 = 0$.We can rewrite $t^2 - 4t + 2$ as $(t^2 - 4t + 4) - 2 = (t-2)^2 - 2$.So,$|t-2| + (t-2)^2 - 2 = 0$.Let $u = |t-2|$. Then $u^2 + u - 2 = 0$.$(u+2)(u-1) = 0$.Since $u = |t-2| \ge 0$,we must have $u = 1$.$|t-2| = 1$.$t-2 = 1$ or $t-2 = -1$.$t = 3$ or $t = 1$.Since $t = \sqrt{x}$,we have $\sqrt{x} = 3$ or $\sqrt{x} = 1$.$x = 9$ or $x = 1$.The sum of the solutions is $9 + 1 = 10$.

The sum of the solutions of the equation $|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0$ for $x>0$ is equal to

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