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

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(C) Given equation: $|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0$ for $x>0$. Let $t = \sqrt{x}$,where $t>0$. The equation becomes $|t-2|+t(t-4)+2=0$. $|t-2|

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(C) Given equation: $|\sqrt{x}-2|+\sqrt{x}(\sqrt{x}-4)+2=0$ for $x>0$. 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,$|t-2|+(t-2)^2-2=0$. Since $(t-2)^2 = |t-2|^2$,we have $|t-2|^2+|t-2|-2=0$. Let $u = |t-2|$,where $u \ge 0$. $u^2+u-2=0$. $(u+2)(u-1)=0$. Since $u \ge 0$,we have $u=1$. $|t-2|=1 \implies t-2=1$ 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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