The sum of the squares of all the roots of th | vedclass

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(C) Given equation: $x^2+|2x-3|-4=0$.Case $I$: $x \geq \frac{3}{2}$.The equation becomes $x^2 + 2x - 3 - 4 = 0$,which is $x^2 + 2x - 7 = 0$.Using the

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$15 - 4\sqrt{2}$

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(C) Given equation: $x^2+|2x-3|-4=0$.Case $I$: $x \geq \frac{3}{2}$.The equation becomes $x^2 + 2x - 3 - 4 = 0$,which is $x^2 + 2x - 7 = 0$.Using the quadratic formula,$x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.Since $x \geq \frac{3}{2}$,we accept $x = 2\sqrt{2} - 1$.Case $II$: $x The equation becomes $x^2 - (2x - 3) - 4 = 0$,which is $x^2 - 2x - 1 = 0$.Using the quadratic formula,$x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.Since $x The roots are $2\sqrt{2}-1$,$1+\sqrt{2}$,and $1-\sqrt{2}$.Sum of squares $= (2\sqrt{2}-1)^2 + (1+\sqrt{2})^2 + (1-\sqrt{2})^2$.$= (8 - 4\sqrt{2} + 1) + (1 + 2\sqrt{2} + 2) + (1 - 2\sqrt{2} + 2)$.$= 9 - 4\sqrt{2} + 3 + 2\sqrt{2} + 3 - 2\sqrt{2} = 15 - 4\sqrt{2}$.

The sum of the squares of all the roots of the equation $x^2+|2x-3|-4=0$ is:

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