The sum of the squares of all the roots of the equation $x^{2}+|2x-3|-4=0$ is:
Step-by-Step Solution
Key Concept: Split into cases based on the sign of $2x-3$. Within each case, solve the resulting quadratic and discard roots that violate the case's hypothesis.
\textbf{Case 1: } $x\ge \dfrac{3}{2}$. Then $|2x-3|=2x-3$, giving
$$x^{2}+2x-7=0\ \Longrightarrow\ x = -1\pm 2\sqrt{2}.$$
Only $x_1 = -1+2\sqrt{2}\approx 1.828\ge 3/2$ is admissible.
\textbf{Case 2: } $x<\dfrac{3}{2}$. Then $|2x-3|=3-2x$, giving
$$x^{2}-2x-1=0\ \Longrightarrow\ x = 1\pm\sqrt{2}.$$
Only $x_2 = 1-\sqrt{2}\approx -0.414 < 3/2$ is admissible.
Sum of squares:
$$x_1^{2}+x_2^{2} = (2\sqrt{2}-1)^{2}+(1-\sqrt{2})^{2} = (9-4\sqrt{2})+(3-2\sqrt{2}) = 12-6\sqrt{2} = 6(2-\sqrt{2}).$$
Correct Answer: 3