If the difference between the roots of the eq | vedclass

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

Question

(D) Let $\alpha, \beta$ be the roots of $x^2+ax+b=0$ and $\gamma, \delta$ be the roots of $x^2+bx+a=0$. From the relations between roots and coefficie

Vedclass · Accepted Answer

$a+b+4=0$

Vedclass · Answer

(D) Let $\alpha, \beta$ be the roots of $x^2+ax+b=0$ and $\gamma, \delta$ be the roots of $x^2+bx+a=0$. From the relations between roots and coefficients: $\alpha+\beta = -a, \alpha\beta = b$ $\gamma+\delta = -b, \gamma\delta = a$ Given that the difference between the roots is the same: $|\alpha-\beta| = |\gamma-\delta|$ Squaring both sides: $(\alpha-\beta)^2 = (\gamma-\delta)^2$ $(\alpha+\beta)^2 - 4\alpha\beta = (\gamma+\delta)^2 - 4\gamma\delta$ $(-a)^2 - 4b = (-b)^2 - 4a$ $a^2 - 4b = b^2 - 4a$ $a^2 - b^2 + 4a - 4b = 0$ $(a-b)(a+b) + 4(a-b) = 0$ $(a-b)(a+b+4) = 0$ Since $a 
eq b$,we have $a-b 
eq 0$. Therefore,$a+b+4 = 0$.

If the difference between the roots of the equations $x^2+ax+b=0$ and $x^2+bx+a=0$ is the same,and $a \neq b$,then:

Similar Questions

The condition that the roots of $x^3-b x^2+c x-d=0$ are in geometric progression is

The harmonic mean of two numbers is $-\frac{8}{5}$ and their geometric mean is $2$. The quadratic equation whose roots are twice those numbers is

If $\alpha$ and $\beta$ are the roots of the equation $4x^2 + 3x + 7 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} = $

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-6x^2+11x-6=0$ and if $a=\alpha^2+\beta^2+\gamma^2$,$b=\alpha\beta+\beta\gamma+\gamma\alpha$ and $c=(\alpha+\beta)(\beta+\gamma)(\gamma+\alpha)$,then the correct inequality among the following is

The number of integers $k$ for which the equation $x^3-27x+k=0$ has at least two distinct integer roots is

Vedclass Test Series

Exam Paper Generator

Online Exam Module