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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the equation: $x^2 + |2x - 3| - 4 = 0$.Case $1$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) Given the equation: $x^2 + |2x - 3| - 4 = 0$.Case $1$: If $x \ge \frac{3}{2}$,then $|2x - 3| = 2x - 3$.$x^2 + 2x - 3 - 4 = 0 \implies x^2 + 2x - 7 = 0$.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 $x \ge \frac{3}{2}$,we check the values: $-1 + 2\sqrt{2} \approx -1 + 2(1.414) = 1.828 > 1.5$ (Valid).$-1 - 2\sqrt{2} So,$x_1 = 2\sqrt{2} - 1$.Case $2$: If $x $x^2 - 2x + 3 - 4 = 0 \implies x^2 - 2x - 1 = 0$.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 $x  1.5$ (Invalid).$1 - \sqrt{2} \approx -0.414 So,$x_2 = 1 - \sqrt{2}$.Sum of the roots: $x_1 + x_2 = (2\sqrt{2} - 1) + (1 - \sqrt{2}) = \sqrt{2}$.

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

Similar Questions

If $x^2-5x+6$ is a factor of $f(x)=x^4-17x^3+kx^2-247x+210$,then the other quadratic factor of $f(x)$ is

If the product of roots of the equation ${x^2} - 3kx + 2{e^{2\log k}} - 1 = 0$ is $7$,then its roots will be real when:

The number of pairs of consecutive positive even integers such that the sum of their squares is $290$ is

Solve the equation $27 x^{2}-10 x+1=0$.

If the equation $x^2 + \lambda x + \mu = 0$ has equal roots and one root of the equation $x^2 + \lambda x - 12 = 0$ is $2$,then $(\lambda, \mu) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module