The sum of the squares of the roots of |x-2|^ | vedclass

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(B) For the equation $|x-2|^2+|x-2|-2=0$:Let $t = |x-2|$,then $t^2+t-2=0$.$(t+2)(t-1)=0 \Rightarrow t=1$ (since $t \geq 0$).$|x-2|=1 \Rightarrow x-2=1

Vedclass · Accepted Answer

$36$

Vedclass · Answer

(B) For the equation $|x-2|^2+|x-2|-2=0$:Let $t = |x-2|$,then $t^2+t-2=0$.$(t+2)(t-1)=0 \Rightarrow t=1$ (since $t \geq 0$).$|x-2|=1 \Rightarrow x-2=1$ or $x-2=-1$.$x=3, 1$.Sum of squares of roots $= 3^2+1^2 = 9+1 = 10$.For the equation $x^2-2|x-3|-5=0$:Case-$I$: $x \geq 3$,then $x^2-2(x-3)-5=0$ $\Rightarrow x^2-2x+1=0$ $\Rightarrow (x-1)^2=0$ $\Rightarrow x=1$.Since $1 Case-$II$: $x Roots $\alpha, \beta = \frac{-2 \pm \sqrt{4 - 4(1)(-11)}}{2} = -1 \pm 2\sqrt{3}$.Both roots are less than $3$,so both are valid.Sum of squares of roots $= (\alpha+\beta)^2 - 2\alpha\beta = (-2)^2 - 2(-11) = 4 + 22 = 26$.Total sum $= 10 + 26 = 36$.

The sum of the squares of the roots of $|x-2|^2+|x-2|-2=0$ and the squares of the roots of $x^2-2|x-3|-5=0$ is:

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