[JEE Main 2014] let alpha beta be the roots of px 2 qx r 0 p 0
Let $\alpha,\beta$ be the roots of $px^2+qx+r=0$ ($p\ne0$). If $p,q,r$ are in A.P. and $\frac1\alpha+\frac1\beta=4$, then $|\alpha-\beta|$ is
(a) $\frac{\sqrt{34}}9$
(b) $\frac{2\sqrt{13}}9$
(c) $\frac{\sqrt{61}}9$
(d) $\frac{2\sqrt{17}}9$
1 Answer
VAVidaara Admin
✓ Vidaara Team
✓ Accepted
· 2d ago
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Correct answer: (b) $\frac{2\sqrt{13}}9$
$-\frac qr=4\Rightarrow q=-4r$; A.P. $2q=p+r\Rightarrow p=-9r$. Then $\alpha+\beta=-\frac49,\alpha\beta=-\frac19$, so $(\alpha-\beta)^2=\frac{52}{81}$, $|\alpha-\beta|=\frac{2\sqrt{13}}9$.
JEE Main 2014 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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