The sum of the roots of the equation,{x^2} + | vedclass

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(C) Given equation: ${x^2} + |2x - 3| - 4 = 0$Case $I$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.The equation becomes: ${x^2} + 2x - 3 - 4 = 0

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$\sqrt{2}$

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(C) Given equation: ${x^2} + |2x - 3| - 4 = 0$Case $I$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.The equation becomes: ${x^2} + 2x - 3 - 4 = 0 \Rightarrow {x^2} + 2x - 7 = 0$.Roots are $x = \frac{-2 \pm \sqrt{4 - 4(1)(-7)}}{2} = \frac{-2 \pm \sqrt{32}}{2} = -1 \pm 2\sqrt{2}$.Since $x \ge \frac{3}{2}$,we check the values: $-1 + 2\sqrt{2} \approx -1 + 2.828 = 1.828 > 1.5$ (Valid).$-1 - 2\sqrt{2} \approx -3.828 So,$x_1 = 2\sqrt{2} - 1$.Case $II$: If $x The equation becomes: ${x^2} - 2x + 3 - 4 = 0 \Rightarrow {x^2} - 2x - 1 = 0$.Roots are $x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$.Since $x  1.5$ (Invalid).$1 - \sqrt{2} \approx -0.414 So,$x_2 = 1 - \sqrt{2}$.The sum of the valid roots is $(2\sqrt{2} - 1) + (1 - \sqrt{2}) = \sqrt{2}$.

The sum of the roots of the equation,${x^2} + |2x - 3| - 4 = 0$,is

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