(B) Let the roots of the equation $px^2+qx+r=0$ be $a\alpha$ and $b\alpha$. From the relation between roots and coefficients: Sum of roots: $a\alpha +

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(B) Let the roots of the equation $px^2+qx+r=0$ be $a\alpha$ and $b\alpha$. From the relation between roots and coefficients: Sum of roots: $a\alpha + b\alpha = (a+b)\alpha = -\frac{q}{p}$ Product of roots: $(a\alpha)(b\alpha) = ab\alpha^2 = \frac{r}{p}$ Now, consider the ratio: $\frac{ab\alpha^2}{(a+b)^2\alpha^2} = \frac{ab}{(a+b)^2}$ Substituting the values: $\frac{ab}{(a+b)^2} = \frac{r/p}{(-q/p)^2} = \frac{r}{p} \times \frac{p^2}{q^2} = \frac{pr}{q^2}$

Class 11 Mathematics 2 Complex Numbers and Quadratic Equations: Quadratic Equations Relation between roots and coefficients

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If the ratio of the roots of the equation $px^2+qx+r=0$ is $a:b$, then $\frac{ab}{(a+b)^2}=$

WBJEE 2011 All WBJEE

A
$\frac{p^2}{qr}$
B
$\frac{pr}{q^2}$
C
$\frac{q^2}{pr}$
D
$\frac{pq}{r^2}$

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2 Complex Numbers and Quadratic Equations: Quadratic Equations

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Relation between roots and coefficients

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