Quadratic Equations
Quadratic Equation
nta_pyq_2025_jan
Grade 11
Question:
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 $X=\{x\in\mathbb{Z}\,:\,\alpha<x<\beta\}$, then $\displaystyle\sum_{x\in X}x^{2}$ is equal to:
Step-by-Step Solution
Key Concept: `No real root' $\Leftrightarrow$ discriminant $<0$. This yields a quadratic inequality in $a$. The resulting open interval $(\alpha,\beta)$ then defines a finite set of integers $X$, and $\sum x^{2}$ uses the identity $\sum_{1}^{n}k^{2}=\dfrac{n(n+1)(2n+1)}{6}$.
Rewrite: $2x^{2}+(a-5)x+(15-3a)=0$. Discriminant
$$D = (a-5)^{2}-8(15-3a) = a^{2}+14a-95.$$
No real root $\Leftrightarrow D<0 \Leftrightarrow (a+19)(a-5)<0\Leftrightarrow a\in(-19,5).$ So $\alpha=-19,\ \beta=5$.
Thus $X=\{-18,-17,\dots,3,4\}$. Compute
$$\sum_{x\in X}x^{2} = \sum_{k=1}^{18}k^{2} + \sum_{k=1}^{4}k^{2} = \frac{18\cdot 19\cdot 37}{6} + \frac{4\cdot 5\cdot 9}{6} = 2109+30 = 2139.$$
Correct Answer: 4