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Mathematics · 2026

JEE Main · 21 January 2026, Shift 2 · Q4

Let α and β be the roots of the equation x^2+2 a x+(3 a +10)=0 such that α<1<β. Then the set of all possible values of a is:

Let $\displaystyle \alpha$ and $\displaystyle \beta$ be the roots of the equation $\displaystyle x^2+2 \mathrm{a} x+(3 \mathrm{a}+10)=0$ such that $\displaystyle \alpha<1<\beta$. Then the set of all possible values of a is :
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JEE Main 2026 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.