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If $r$ is a constant such that $|x^{2} – 4x -13| = r$ has exactly three distinct real roots, then the value of $r$ is

  1. $15$
  2. $18$
  3. $17$
  4. $21$
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Given that,  $|x^{2}-4x-13| = r\quad \longrightarrow (1)$

We know that, $|x| = \left\{\begin{matrix}x \;;&x \geq 0 \\ -x\;; &x<0 \end{matrix}\right.$

$\textbf{Case 1:}\; x^{2}-4x-13 = r$

$\Rightarrow x^{2}-4x-13-r = 0$

For two distinct real roots, the discriminant will be greater than zero.

$\Rightarrow D_{1}>0$

$\Rightarrow b^{2}-4ac>0 \quad [\because \text{For the quadratic equation}\; ax^{2} + bx +c = 0]$

$\Rightarrow (-4)^{2}-4(-13-r)>0$

$\Rightarrow 16+52+4r>0$

$\Rightarrow 4r+68>0$

$\Rightarrow \boxed{r>-17}$

$\textbf{Case 2:}\; x^{2}-4x-13 = -r$

$\Rightarrow x^{2}-4x-13+r = 0$

For one real root, the discriminant will be zero.

$\Rightarrow D_{2} =0$

$\Rightarrow b^{2}-4ac=0$

$\Rightarrow (-4)^{2}-4(1)(-13+r)=0$

$\Rightarrow 16+52-4r=0$

$\Rightarrow 68-4r=0$

$\Rightarrow 4r=68$

$\Rightarrow \boxed{r=17}$

$\therefore$ The value of $r = 17$

Correct Answer $:\text{C}$


$\textbf{PS:}$  

We know the quadratic formula is

$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

for any quadratic equation written in standard form of $ax^2+bx+c=0.$ The discriminant $D$ for the quadratic equation is

$$D=b^2-4ac,$$

where

$$\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}​$$

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