If the Set of All a ∈ R for Which 2x² + (a−5)x + 15 = 3a Has No Real Root Is the Interval (α, β), and X = {x ∈ Z : x Lies Between α and β}, Then the Sum of x² Over X Is Equal To

JEE Main 2025 — 29 January, Shift 2. Previous Year Question.

Problem

If the set of all $a\in\mathbb R$, for which the equation $2x^2+(a-5)x+15=3a$ has no real root, is the interval $(\alpha,\beta)$, and $X=\{x\in\mathbb Z: x \text{ lies between } \alpha \text{ and } \beta\}$, then $\sum_{x\in X}x^2$ is equal to:

Key insight. “No real root” translates directly into a discriminant-less-than-zero inequality, and that inequality — a quadratic in $a$ itself — has to be solved before the problem’s actual question about integers can even be asked.

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Approach

First rewrite the given equation in standard quadratic form $Ax^2+Bx+C=0$, then apply the no-real-root condition $B^2-4AC<0$ — this produces a quadratic inequality in $a$ whose solution set is exactly the interval $(\alpha,\beta)$. Once $\alpha$ and $\beta$ are known, list every integer strictly between them and sum their squares using the standard formula for $\sum k^2$, adjusted for the fact that the integers run through both negative and positive values.

Solution

Step 1 — Rewrite the equation in standard form

$$2x^2+(a-5)x+15=3a \implies 2x^2+(a-5)x+(15-3a) = 0$$

Here $A=2$, $B=a-5$, $C=15-3a$.

Step 2 — Apply the no-real-root condition

A quadratic has no real root exactly when its discriminant is negative:

$$B^2-4AC < 0 \implies (a-5)^2-4(2)(15-3a) < 0$$

Expanding:

$$a^2-10a+25-120+24a < 0 \implies a^2+14a-95 < 0$$

Step 3 — Solve the quadratic inequality in a

Find the roots of $a^2+14a-95=0$ using the quadratic formula:

$$a = \frac{-14\pm\sqrt{14^2-4(1)(-95)}}2 = \frac{-14\pm\sqrt{196+380}}2 = \frac{-14\pm\sqrt{576}}2 = \frac{-14\pm24}2$$

$$a = \frac{10}2=5 \quad \text{or} \quad a=\frac{-38}2=-19$$

Since the coefficient of $a^2$ is positive, the parabola in $a$ opens upward, so $a^2+14a-95<0$ holds between the roots:

$$-19 < a < 5$$

So $\alpha=-19$ and $\beta=5$.

Step 4 — List the integers strictly between α and β

$$X = \{x\in\mathbb Z: -19<x<5\} = \{-18,-17,\ldots,-1,0,1,2,3,4\}$$

This set has $4-(-18)+1=23$ integers.

Step 5 — Sum the squares using symmetry

Since $x^2=(-x)^2$, the squares from $-18$ to $-1$ match the squares from $1$ to $18$:

$$\sum_{x=-18}^{4}x^2 = \sum_{x=1}^{18}x^2+0^2+\sum_{x=1}^{4}x^2$$

Using $\displaystyle\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6$:

$$\sum_{x=1}^{18}x^2 = \frac{18\times19\times37}6 = \frac{12654}6 = 2109$$

$$\sum_{x=1}^{4}x^2 = \frac{4\times5\times9}6 = \frac{180}6 = 30$$

Step 6 — Add the two parts

$$\sum_{x\in X}x^2 = 2109+30 = 2139$$

Answer

Common mistakes

• Using ≤ instead of < when applying the discriminant condition. “No real root” strictly means the discriminant is negative, not zero — a discriminant of exactly zero gives one repeated real root, which doesn’t satisfy “no real root.”
• Getting the direction of the inequality backwards. Since the parabola $a^2+14a-95$ opens upward, it’s negative between its roots and positive outside them — mixing this up gives the interval $(-\infty,-19)\cup(5,\infty)$ instead of $(-19,5)$.
• Including or excluding the wrong endpoints when listing X. Because the interval is open, $-19$ and $5$ themselves are not included in $X$ — it’s easy to accidentally include $-19$ or $5$, or to miscount and stop the list one integer too early.
• Forgetting the symmetry shortcut and adding 23 separate squares by hand. Pairing $x^2=(-x)^2$ as in Step 5 turns the sum into two applications of the standard $\sum k^2$ formula, which is far less error-prone than summing 23 terms individually.

Practise next

• If the set of all $a\in\mathbb R$ for which $x^2+(a-3)x+7=2a$ has no real root is $(\alpha,\beta)$, find $\sum_{x\in X}x^2$ for $X=\{x\in\mathbb Z: \alpha<x<\beta\}$, using the same discriminant-then-sum approach.