The sum, of the squares of all the roots of the equation x2 + |2x – 3| – 4 = 0, is

x2+|2x−3|−4=0 Case I:x≥32 x2+2x−3−4=0x2+2x−7=0x==22−1 Case II : x<32x2+3−2x−4=0x2−2x−1=0x=1−2 Sum of squares =(22−1)2+(1−2)2=8−42+1+1−22+2=6(2−2)

The sum, of the squares of all the roots of the equation x2 + |2x – 3| – 4 = 0, is

x2+|2x−3|−4=0 Case I:x≥32 x2+2x−3−4=0x2+2x−7=0x==22−1 Case II : x<32x2+3−2x−4=0x2−2x−1=0x=1−2 Sum of squares =(22−1)2+(1−2)2=8−42+1+1−22+2=6(2−2)

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The sum, of the squares of all the roots of the equation x² + |2x – 3| – 4 = 0, is

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$6 (3 - \sqrt{2})$

$6 (2 - \sqrt{2})$

$3 (2 - \sqrt{2})$

$3 (3 - \sqrt{2})$

answer is C.

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Detailed Solution

$\begin{array}{l} x^{2} + | 2 x - 3 | - 4 = 0 \\ Case I : x \geq \frac{3}{2} \\ x^{2} + 2 x - 3 - 4 = 0 \end{array}$

$\begin{array}{l} \begin{matrix} x^{2} + 2 x - 7 = 0 \\ x = = 2 \sqrt{2} - 1 \end{matrix} \\ Case II : x < \frac{3}{2} \\ x^{2} + 3 - 2 x - 4 = 0 \\ x^{2} - 2 x - 1 = 0 \\ x = 1 - \sqrt{2} \\ Sum of squares = (2 \sqrt{2} - 1)^{2} + (1 - \sqrt{2})^{2} \\ = 8 - 4 \sqrt{2} + 1 + 1 - 2 \sqrt{2} + 2 \\ = 6 (2 - \sqrt{2}) \end{array}$