Find the condition to be satisfied by the coefficients of the equation $p x^{2} + q x + r = 0$ , so that the roots are in the ratio 3:4.

12 q^{2} = 49 p r

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12 q^{2} = - 49 p r

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49 q^{2} = 12 p r

No worries! We‘ve got your back. Try BYJU‘S free classes today!

49 q^{2} = - 12 p r

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Solution

The correct option is D
$12 q^{2} = 49 p r$

Let the common factor for the roots be y.
The roots are 3y and 4y, $3 y + 4 y = \frac{- q}{p}$

$y = \frac{- q}{7 p}$

$(3 y) (4 y) = \frac{r}{p}$

$y^{2} = \frac{r}{12 p}$

$y^{2} = (- \frac{q}{7 p})^{2} = \frac{r}{12 p}$

$12 q^{2} = 49 p r$

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