If the set of all a \in R , for which the equation 2 x^2+(a-5) x+15=3 a has no real root, is the interval ( \alpha, \beta ), and X=|x \in Z ; \alpha \lt x \lt \beta| , then \sum\limits_x \in X x^2 is equal to:

Quadratic Equation and Inequalities PYQ — JEE Main 2025 · 29 Jan · Shift 2 · Q32 with Solution

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Correct answer: A

Given $2x^2+(a-5)x+15=3a$ bring all terms to one side: $2x^2+(a-5)x+(15-3a)=0$

This is a quadratic in $x$ with:

\quad B=a-5, \quad C=15-3a$$ 2. **Condition for no real root** For a quadratic equation in $x$ to have no real root, its discriminant must be negative: $$D=B^2-4AC<0$$ So, $$D=(a-5)^2-4(2)(15-3a)<0$$ Now simplify: $$(a-5)^2-8(15-3a)<0$$ $$a^2-10a+25-120+24a<0$$ $$a^2+14a-95<0$$ 3. **Solve the inequality in $a$** We solve: $$a^2+14a-95<0$$ Find the roots: $$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}$$ Thus roots are: $$a=5,\,-19$$ Since the coefficient of $a^2$ is positive, the quadratic is negative between the roots. Hence, $$(\alpha,\beta)=(-19,5)$$ So, $$\alpha=-19,\quad \beta=5$$ 4. **Find the set $X$** Given $$X=\{x\in \mathbb Z:\alpha<x<\beta\}$$ that is, $$-19<x<5$$ So the integers in $X$ are: $$-18,-17,-16,\dots,-1,0,1,2,3,4$$ 5. **Compute $\sum_{x\in X} x^2$** We need: $$\sum_{x=-18}^{4} x^2$$ Break it as: $$\sum_{x=-18}^{-1}x^2+0^2+\sum_{x=1}^{4}x^2$$ Since squaring removes sign, $$=\sum_{k=1}^{18}k^2+\sum_{k=1}^{4}k^2$$ Now, $$\sum_{k=1}^{18}k^2=\frac{18\cdot19\cdot37}{6}$$ $$=3\cdot19\cdot37=2109$$ Also, $$\sum_{k=1}^{4}k^2=1^2+2^2+3^2+4^2=1+4+9+16=30$$ Therefore, $$\sum_{x\in X}x^2=2109+30=2139$$ 6. **Match with the options** Thus the required value is: $$\boxed{2139}$$ So the correct option is **A**.

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Correct answer: A

Given $2x^2+(a-5)x+15=3a$ bring all terms to one side: $2x^2+(a-5)x+(15-3a)=0$

This is a quadratic in $x$ with:

\quad B=a-5, \quad C=15-3a$$ 2. **Condition for no real root** For a quadratic equation in $x$ to have no real root, its discriminant must be negative: $$D=B^2-4AC<0$$ So, $$D=(a-5)^2-4(2)(15-3a)<0$$ Now simplify: $$(a-5)^2-8(15-3a)<0$$ $$a^2-10a+25-120+24a<0$$ $$a^2+14a-95<0$$ 3. **Solve the inequality in $a$** We solve: $$a^2+14a-95<0$$ Find the roots: $$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}$$ Thus roots are: $$a=5,\,-19$$ Since the coefficient of $a^2$ is positive, the quadratic is negative between the roots. Hence, $$(\alpha,\beta)=(-19,5)$$ So, $$\alpha=-19,\quad \beta=5$$ 4. **Find the set $X$** Given $$X=\{x\in \mathbb Z:\alpha<x<\beta\}$$ that is, $$-19<x<5$$ So the integers in $X$ are: $$-18,-17,-16,\dots,-1,0,1,2,3,4$$ 5. **Compute $\sum_{x\in X} x^2$** We need: $$\sum_{x=-18}^{4} x^2$$ Break it as: $$\sum_{x=-18}^{-1}x^2+0^2+\sum_{x=1}^{4}x^2$$ Since squaring removes sign, $$=\sum_{k=1}^{18}k^2+\sum_{k=1}^{4}k^2$$ Now, $$\sum_{k=1}^{18}k^2=\frac{18\cdot19\cdot37}{6}$$ $$=3\cdot19\cdot37=2109$$ Also, $$\sum_{k=1}^{4}k^2=1^2+2^2+3^2+4^2=1+4+9+16=30$$ Therefore, $$\sum_{x\in X}x^2=2109+30=2139$$ 6. **Match with the options** Thus the required value is: $$\boxed{2139}$$ So the correct option is **A**.