Quadratic Equations
Roots with α<1<β — Range of a
nta_pyq_2026_jan
Grade 11
Question:
Let $\alpha$ and $\beta$ be the roots of the equation $x^2+2ax+(3a+10)=0$ such that $\alpha<1<\beta$. Then the set of all possible values of $a$ is:
$\left(-\infty,-\dfrac{11}{5}\right)$
$(-\infty,-2)\cup(5,\infty)$
$(-\infty,-3)$
$\left(-\infty,-\dfrac{11}{5}\right)\cup(5,\infty)$
Step-by-Step Solution
Key Concept: For $\alpha<1<\beta$ with positive leading coefficient, we need $f(1)<0$. $f(1)=1+2a+3a+10=5a+11<0\Rightarrow a<-\tfrac{11}{5}$.
$a\in\left(-\infty,-\dfrac{11}{5}\right)$.
Correct Answer: 1