If alpha and beta are the roots of the equati | vedclass

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

Question

(B) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $a\alpha^2 + b\alpha + c = 0$ and $a\beta^2 + b\beta + c = 0$.From $a\alpha^2

Vedclass · Accepted Answer

$\frac{b^2 - 2ac}{a^2c^2}$

Vedclass · Answer

(B) Since $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $a\alpha^2 + b\alpha + c = 0$ and $a\beta^2 + b\beta + c = 0$.From $a\alpha^2 + b\alpha + c = 0$,we get $\alpha(a\alpha + b) = -c$,so $(a\alpha + b) = -\frac{c}{\alpha}$.Similarly,$(a\beta + b) = -\frac{c}{\beta}$.Substituting these into the expression:$\frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2} = \frac{1}{(-c/\alpha)^2} + \frac{1}{(-c/\beta)^2} = \frac{\alpha^2}{c^2} + \frac{\beta^2}{c^2} = \frac{\alpha^2 + \beta^2}{c^2}$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$,where $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$:$\frac{(\alpha + \beta)^2 - 2\alpha\beta}{c^2} = \frac{(-b/a)^2 - 2(c/a)}{c^2} = \frac{b^2/a^2 - 2c/a}{c^2} = \frac{b^2 - 2ac}{a^2c^2}$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then what is the value of $\frac{1}{(a\alpha + b)^2} + \frac{1}{(a\beta + b)^2}$?

Similar Questions

If $\alpha, \beta$ are the roots of $x^2-5 \gamma x-6 \delta=0$ and $\gamma, \delta$ are the roots of $x^2-5 \alpha x-6 \beta=0$,then $\alpha+\beta+\gamma+\delta=$

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

If one root of the equation $ax^2 + bx + c = 0$ is the square of the other,then $a(c - b)^3 = cX$,where $X$ is

If the roots of the equation $x^2 + bx + c = 0$ are reciprocals of each other,then...

If $2$ and $6$ are the roots of the equation $ax^2 + bx + 1 = 0$,then the quadratic equation,whose roots are $\frac{1}{2a + b}$ and $\frac{1}{6a + b}$,is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module