JEE Main 2025 — 28 January, Morning Shift. Previous Year Question.
Problem
The sum of the squares of all the roots of the equation $x^2+|2x-3|-4=0$ is:
Key insight. $|2x-3|$ switches its formula at $x=\dfrac32$ — so the equation genuinely splits into two different quadratics on either side of that point. Each quadratic has two roots on paper, but only the ones that actually land in their own case’s domain are real solutions of the original equation — checking this is not optional bookkeeping, it’s what determines the answer.
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Approach
Split $|2x-3|$ at its sign-change point $x=\dfrac32$ into two cases. In each case, solve the resulting quadratic and keep only the root(s) that satisfy that case’s own domain restriction. Once the two valid roots (one from each case) are found, square and add them.
Solution
Step 1 — Case 1: x ≥ 3/2, where |2x−3| = 2x−3
$$x^2+(2x-3)-4=0 \implies x^2+2x-7=0$$
$$x = \frac{-2\pm\sqrt{4+28}}{2} = \frac{-2\pm\sqrt{32}}{2} = -1\pm2\sqrt2$$
Step 2 — Check which root from Case 1 satisfies x ≥ 3/2
$-1+2\sqrt2 \approx -1+2.83 = 1.83 \geq 1.5$ ✓ — valid.
$-1-2\sqrt2 \approx -3.83$, not $\geq1.5$ — rejected.
So $\alpha = -1+2\sqrt2$.
Step 3 — Case 2: x < 3/2, where |2x−3| = 3−2x
$$x^2+(3-2x)-4=0 \implies x^2-2x-1=0$$
$$x = \frac{2\pm\sqrt{4+4}}{2} = \frac{2\pm\sqrt8}{2} = 1\pm\sqrt2$$
Step 4 — Check which root from Case 2 satisfies x < 3/2
$1+\sqrt2\approx2.41$, not $<1.5$ — rejected.
$1-\sqrt2\approx-0.41 < 1.5$ ✓ — valid.
So $\beta = 1-\sqrt2$.
Step 5 — Compute α² and β²
$$\alpha^2 = (-1+2\sqrt2)^2 = 1-4\sqrt2+8 = 9-4\sqrt2$$
$$\beta^2 = (1-\sqrt2)^2 = 1-2\sqrt2+2 = 3-2\sqrt2$$
Step 6 — Add them
$$\alpha^2+\beta^2 = (9-4\sqrt2)+(3-2\sqrt2) = 12-6\sqrt2 = 6(2-\sqrt2)$$
Answer
$$6(2-\sqrt2)$$
Common mistakes
- Keeping both roots from each case without checking the domain. Each quadratic naturally produces two roots algebraically, but only one root per case actually satisfies the domain condition that generated that case — including the rejected roots would give a wrong, larger sum.
- Using the wrong sign for $|2x-3|$ in each case. For $x\geq\dfrac32$, $2x-3\geq0$ so $|2x-3|=2x-3$; for $x<\dfrac32$, $2x-3<0$ so $|2x-3|=-(2x-3)=3-2x$ — swapping these flips the resulting quadratic entirely.
Practise next
- Find the sum of the squares of all the roots of $x^2+|2x-1|-3=0$, using the same case-split-and-domain-check approach.
Show answer
$10-2\sqrt5-2\sqrt3\approx2.064$. Split at $x=\tfrac12$, where $2x-1$ changes sign.
For $x\ge\tfrac12$: $x^2+2x-4=0$, so $x=-1\pm\sqrt5$; only $\sqrt5-1\approx1.24$ lies in the case’s domain. For $x<\tfrac12$: $x^2-2x-2=0$, so $x=1\pm\sqrt3$; only $1-\sqrt3\approx-0.73$ does.
So the roots are $\sqrt5-1$ and $1-\sqrt3$, and their squares sum to $(6-2\sqrt5)+(4-2\sqrt3)$. The domain check is what removes the other two candidates — skipping it gives four ‘roots’.
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