Modulus Equations PYQ | JEE Main 2025 | Root Sum Trick

 

❓ Question

Find the sum of the squares of the roots of the equations

$$|x-2|^2+|x-2|-2=0$$

and

$$x^2-2|x-3|-5=0.$$

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✍️ Solution

Equation 1

$$|x-2|^2+|x-2|-2=0.$$

Let

$$y=|x-2|.$$

Then,

$$y^2+y-2=0.$$

Factorizing,

$$(y+2)(y-1)=0.$$

Since

$$y=|x-2|\ge0,$$ $$y=1.$$

Thus,

$$|x-2|=1.$$

So,

$$x-2=\pm1.$$

Hence,

$$x=3,\,-1.$$

Therefore,

$$\text{Sum of squares of roots} = 3^2+(-1)^2 = 9+1 = 10.$$

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Equation 2

$$x^2-2|x-3|-5=0.$$

Solve case-wise.

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Case 1: $x\ge3$

$$|x-3|=x-3.$$

The equation becomes

$$x^2-2(x-3)-5=0.$$
$$x^2-2x+1=0.$$
$$(x-1)^2=0.$$

This gives

$$x=1,$$

which does not satisfy $x\ge3$.

No valid solution.

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Case 2: $x<3$

$$|x-3|=3-x.$$

The equation becomes

$$x^2-2(3-x)-5=0.$$
$$x^2+2x-11=0.$$

Using the quadratic formula,

$$x = \frac{-2\pm\sqrt{4+44}}{2} = -1\pm2\sqrt3.$$

Both roots satisfy

$$x<3.$$

Therefore,

$$\text{Sum of squares of roots} = (-1-2\sqrt3)^2+(-1+2\sqrt3)^2.$$

Using

$$(a-b)^2+(a+b)^2=2(a^2+b^2),$$

with

$$a=-1,\qquad b=2\sqrt3,$$

we get

$$2(1+12)=26.$$

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Total Sum

$$10+26=36.$$

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✅ Final Answer

$$\boxed{36}$$