For the equation 2x^2 + 2(a + b)x + a^2 + b^2 | vedclass

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

Question

(B) Given the equation $2x^2 + 2(a + b)x + a^2 + b^2 = 0$. Sum of roots $\alpha + \beta = -\frac{2(a + b)}{2} = -(a + b)$. Product of roots $\alpha\be

Vedclass · Accepted Answer

$x^2 - 4abx - (a^2 - b^2)^2 = 0$

Vedclass · Answer

(B) Given the equation $2x^2 + 2(a + b)x + a^2 + b^2 = 0$. Sum of roots $\alpha + \beta = -\frac{2(a + b)}{2} = -(a + b)$. Product of roots $\alpha\beta = \frac{a^2 + b^2}{2}$. Now,$(\alpha + \beta)^2 = (-(a + b))^2 = (a + b)^2$. And $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (a + b)^2 - 4(\frac{a^2 + b^2}{2}) = a^2 + 2ab + b^2 - 2a^2 - 2b^2 = -(a^2 - 2ab + b^2) = -(a - b)^2$. The required equation is $x^2 - [(\alpha + \beta)^2 + (\alpha - \beta)^2]x + [(\alpha + \beta)^2(\alpha - \beta)^2] = 0$. Sum of new roots $= (a + b)^2 - (a - b)^2 = 4ab$. Product of new roots $= (a + b)^2 	imes (-(a - b)^2) = -((a + b)(a - b))^2 = -(a^2 - b^2)^2$. Thus,the equation is $x^2 - 4abx - (a^2 - b^2)^2 = 0$.

For the equation $2x^2 + 2(a + b)x + a^2 + b^2 = 0$,if $\alpha$ and $\beta$ are the roots,then the equation whose roots are $(\alpha + \beta)^2$ and $(\alpha - \beta)^2$ is:

Similar Questions

If the roots of the equation $ax^2 + bx + c = 0$ are $\alpha$ and $\beta$,then the roots of the equation $cx^2 + bx + a = 0$ are

If $\alpha, \beta$ and $\gamma$ are the roots of the equation $x^3+3x^2+4x+5=0$,then the cubic equation whose roots are $1+4\alpha$,$1+4\beta$ and $1+4\gamma$ is

If $f(x)=ax^2+bx+c$ satisfies $f(1)+2f(2)=0$ and $2f(1)+f(2)=0$,then $3a+b=$

If the ratio of the roots of the equation $ax^2 + bx + c = 0$ is $p:q$,then

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module