The number of real roots of the equation x | x | - 5 | x + 2 | + 6 = 0 , is [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
The number of real roots of the equation $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0,$ is [2023]

1 5

2 3

3 4

4 6

Ans.
(2)

We have, $x$ $| x |$ $- 5$ $| x + 2 |$ $+ 6 = 0$

Case I: $x \leq - 2$

$- x^{2} + 5 x + 10 + 6 = 0$

$\Rightarrow - x^{2} + 5 x + 16 = 0 \Rightarrow x^{2} - 5 x - 16 = 0$

$\Rightarrow x = \frac{5 \pm \sqrt{{(- 5)}^{2} + 16 \times 4}}{2} = \frac{5 \pm \sqrt{25 + 64}}{2} = \frac{5 \pm \sqrt{89}}{2}$

Only $x = \frac{5 - \sqrt{89}}{2}$ belongs to $(- \infty, - 2]$

So, one solution exists in this case.

Case II: $- 2 < x \leq 0$

$- x^{2} - 5 x - 10 + 6 = 0$

$\Rightarrow - x^{2} - 5 x - 4 = 0 \Rightarrow x^{2} + 5 x + 4 = 0$

$\Rightarrow (x + 1) (x + 4) = 0 \Rightarrow x = - 1 or x = - 4$

Since $- 2 < x \leq 0$ , $x = - 4$ is not a solution.

So, $x = - 1$ is a solution in this case.

Case III: $x > 0$

$x^{2} - 5 x - 4 = 0 \Rightarrow x = \frac{5 \pm \sqrt{25 + 16}}{2} = \frac{5 \pm \sqrt{41}}{2}$

But $x > 0$ . So, $x = \frac{5 + \sqrt{41}}{2}$ is the only solution in this case.

Therefore, the equation has three solutions.

Want Us to Email you About Special Offers & Updates?