If the ratio of the roots of the equation px^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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 +

Vedclass · Accepted Answer

$\frac{pr}{q^2}$

Vedclass · Answer

(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: $\frac{ab}{(a+b)^2} = \frac{r/p}{(-q/p)^2} = \frac{r}{p} 	imes \frac{p^2}{q^2} = \frac{pr}{q^2}$

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

Similar Questions

What is the geometric mean of the roots of the equation $x^2 - 18x + 9 = 0$?

If $1, 2, 3$ and $4$ are the roots of the equation $x^4+ax^3+bx^2+cx+d=0$,then $a+2b+c$ is equal to

For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$,if the product of the roots is zero,what is the sum of the roots?

If $\alpha, \beta$ are the roots of the equation $x^2-2 \sqrt{3} x+4=0$,then $\alpha^6+\beta^6=$

The harmonic mean of the roots of the equation $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module