If one root of $x^2+p x-q^2=0, p$ and $q$ are real, be less than 2 and other be greater than 2, then

Question

If one root of $x^2+p x-q^2=0, p$ and $q$ are real, be less than 2 and other be greater than 2, then, $4+2 p+q^2 > 0$, $4+2 p+q^2 < 0$, $4+2 p-q^2 > 0$, $4+2 p-q^2 < 0$

MARKS App · Accepted Answer

Hint : $\alpha 2$ Let, $\mathrm{p}(\mathrm{x})=\mathrm{x}^2+\mathrm{px}-\mathrm{q}^2$ $\begin{aligned} & \therefore \mathrm{p}(2) < 0 \ & \Rightarrow 2^2+2 p-\mathrm{q}^2 < 0 \ & \Rightarrow 4+2 p-\mathrm{q}^2 < 0 \end{aligned}$

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