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

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

Question

(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

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(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

Similar Questions

The number of integral values of $m$ for which the quadratic expression $(1 + 2m)x^2 - 2(1 + 3m)x + 4(1 + m)$ is always positive for all $x \in R$ is:

If $a+b=1,$ find the value of $a^{3}+b^{3}-ab-(a^{2}-b^{2})^{2}$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{2} = 14641$
$II.$ $y = \sqrt{14641}$

The expression $x^2-x+1$ has:

If $a < 0$,then the inequality $ax^2 - 2x + 4 > 0$ has the solution represented by:

Vedclass Test Series

Exam Paper Generator

Online Exam Module