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Quantitative Aptitude Question on Linear Inequalities

If r is a constant such that ∣x2−4x−13∣=r|x^2-4x-13| = r has exactly three distinct real roots, then the value of r is

A

15

B

18

C

17

D

21

Answer

17

Explanation

Solution

Given the equation ∣x2−4x−13∣=r|x^2 - 4x - 13| = r
The equation x2−4x−13=rx^2 - 4x - 13 = r or x2−4x−13=−rx^2 - 4x - 13 = -r should have exactly three distinct real roots combined.
1. Let's solve for the equation x2−4x−13=rx^2 - 4x - 13 = r
x2−4x−13−r=0x^2 - 4x - 13 - r = 0
Using the discriminant b2−4acb^2 - 4ac of a quadratic equation ax2+bx+cax^2 + bx + c, for the equation to have real roots, b2−4ac≥0b^2 - 4ac \geq 0.
So, for our equation:
16−4(1)(−13−r)≥016 - 4(1)(-13 - r) \geq 0
16+52+4r≥016 + 52 + 4r \geq 0
68+4r≥068 + 4r \geq 0
r≤−17r \leq -17
2. Let's solve for the equation
x2−4x−13=−rx^2 - 4x - 13 = -r
x2−4x−13+r=0x^2 - 4x - 13 + r = 0
Again, using the discriminant:
16−4(1)(−13+r)≥016 - 4(1)(-13 + r) \geq 0
16+52−4r≥016 + 52 - 4r \geq 0
68−4r≥068 - 4r \geq 0
r≤17r \leq 17
Since we need a total of three real roots for both equations combined and considering that for r≤−17r \leq -17 the first equation will give real roots and the second equation will not, the only possible value from the options that can provide exactly 3 real roots is r=17r = 17.

Therefore, the correct option is (C): 17