JEE Main 2025 Apr 8 Shift 2, Mathematics Q6: Some Other Forms of Equations

JEE Main2025Apr 8, Shift 2Mathematics

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

Solution

$∣ x - 2 ∣^{2} + 2∣ x - 2∣ - ∣ x - 2∣ - 2 = 0$

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

$\Rightarrow ∣ x - 2∣ = 1$

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

$∴$ Sum of squares of the roots $= 3^{2} + 1^{2} = 10$

Now solve

$x^{2} - 2∣ x - 3∣ - 5 = 0$

Case I: $x - 3 \geq 0$

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

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

$\Rightarrow x = 1$

But $x \geq 3$ , which is not satisfied.

$\Rightarrow$ No solution.

Case II: $x-3<0$

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

$D > 0 \Rightarrow$ Real and distinct roots.

Let $f (x) = x^{2} + 2 x - 11$ .

$f(3)=9+6-11=4>0,\qquad -\dfrac{b}{2a}=-1<3$

$\Rightarrow$ Both roots are less than $3$ , hence both satisfy $x<3$.

If the roots are $α, β$ , then

$α + β = - 2, α β = - 11$

Sum of squares $= (α + β)^{2} - 2 α β$

$= 4 + 22 = 26$

$∴$ Final sum $= 10 + 26 = 36$

Concept behind this question

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