Question ID: #446
The sum of the squares of all the roots of the equation $x^2 + |2x-3| – 4 = 0$ is:
- (1) $3(3 – \sqrt{2})$
- (2) $6(3 – \sqrt{2})$
- (3) $6(2 – \sqrt{2})$
- (4) $3(2 – \sqrt{2})$
Solution:
Equation: $x^2 + |2x-3| – 4 = 0$.
Case 1: $2x – 3 \ge 0 \Rightarrow x \ge 1.5$
$x^2 + (2x – 3) – 4 = 0$
$x^2 + 2x – 7 = 0$
Roots: $x = \frac{-2 \pm \sqrt{4 + 28}}{2} = -1 \pm \sqrt{8} = -1 \pm 2\sqrt{2}$.
Check constraint $x \ge 1.5$:
$x_1 = -1 + 2\sqrt{2} \approx -1 + 2.828 = 1.828$ (Accepted).
$x_2 = -1 – 2\sqrt{2}$ (Rejected).
Case 2: $2x – 3 < 0 \Rightarrow x < 1.5$ $x^2 - (2x - 3) - 4 = 0$ $x^2 - 2x - 1 = 0$ Roots: $x = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$. Check constraint $x < 1.5$: $x_3 = 1 + \sqrt{2} \approx 2.414$ (Rejected). $x_4 = 1 - \sqrt{2} \approx -0.414$ (Accepted).
Valid roots are $\alpha = -1 + 2\sqrt{2}$ and $\beta = 1 – \sqrt{2}$.
Sum of squares = $\alpha^2 + \beta^2$.
$\alpha^2 = (-1 + 2\sqrt{2})^2 = 1 + 8 – 4\sqrt{2} = 9 – 4\sqrt{2}$.
$\beta^2 = (1 – \sqrt{2})^2 = 1 + 2 – 2\sqrt{2} = 3 – 2\sqrt{2}$.
Total Sum = $(9 – 4\sqrt{2}) + (3 – 2\sqrt{2}) = 12 – 6\sqrt{2}$.
Factor out 6:
Sum = $6(2 – \sqrt{2})$.
Ans. (3)
Was this solution helpful?
YesNo