The sum of the squares of the roots of and the squares of the roots of is [2025]

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 sum of the squares of the roots of $| x - 2 |^{2} + | x - 2 | - 2 = 0$ and the squares of the roots of $x^{2} - 2 | x - 3 | - 5 = 0$ is [2025]

1 24

2 36

3 30

4 26

Ans.
(2)

We have, $| x - 2 |^{2} + | x - 2 | - 2 = 0$

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

$\Rightarrow (| x - 2 | + 2) (| x - 2 | - 1) = 0$

$\Rightarrow | (x - 2) | = 1$ [ $∵ | x - 2 | \neq - 2$ ]

$\Rightarrow x = 2 \pm 1 \Rightarrow x = 3, 1$

Sum of square of roots = 9 + 1 = 10

Now, we have $x^{2} - 2 | x - 3 | - 5 = 0$

Case I : When x – 3 > 0

$\Rightarrow x^{2} - 2 x + 1 = 0 \Rightarrow {(x - 1)}^{2} = 0 \Rightarrow x = 1$

but x > 3 $\Rightarrow x \neq 1$

Case II : When x – 3 < 0

$\Rightarrow x^{2} + 2 x - 11 = 0$

Discriminant, D = 4 + 44 = 48 > 0

$x = \frac{- 2 \pm \sqrt{48}}{2} = \frac{- 2 \pm 4 \sqrt{3}}{2} = - 1 \pm 2 \sqrt{3}$

SInce, x < 3, so both roots are valid.

Sum of squares of roots = ${(- 1 + 2 \sqrt{3})}^{2} + {(- 1 - 2 \sqrt{3})}^{2}$

$= 1 + 12 - 4 \sqrt{3} + 1 + 12 + 4 \sqrt{3} = 26$

$∴$ Required sum = 10 + 26 = 36.

Want Us to Email you About Special Offers & Updates?