First slide

Theory of equations

Question

Sum of the roots of the equation
$x^{2} + | 2 x - 3 | - 4 = 0$ is

Moderate

A

2
B

-2
C

$\sqrt{2}$
D

$- \sqrt{2}$

Solution

Case 1: When $x < 3 / 2$
In this case the equation becomes
$\begin{array}{ll} \Rightarrow x^{2} - (2 x - 3) - 4 = 0 \\ \Rightarrow x^{2} - 2 x + 1 = 2 \\ \Rightarrow (x - 1)^{2} = 2 \Rightarrow x - 1 = \pm \sqrt{2} \\ \Rightarrow x = 1 \pm \sqrt{2} \end{array}$
As $x < 3 / 2$ we take, $x = 1 - \sqrt{2}$
Case 2: When $x \geq 3 / 2$
In this case the equation becomes
$\begin{array}{l} x^{2} + (2 x - 3) - 4 = 0 \\ \Rightarrow (x + 1)^{2} = 8 \Rightarrow x + 1 \pm 2 \sqrt{2} \end{array}$
As $x \geq 3 / 2, x = - 1 + 2 \sqrt{2}$
Sum of the roots is
$(1 - \sqrt{2}) + (- 1 + 2 \sqrt{2}) = \sqrt{2}$ .

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