The condition for the roots of the equation ( | vedclass

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

Question

(A) For a quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = (c^2 - ab)$,$B =

Vedclass · Accepted Answer

$a = 0$

Vedclass · Answer

(A) For a quadratic equation $Ax^2 + Bx + C = 0$ to have equal roots,the discriminant $D = B^2 - 4AC$ must be equal to $0$.Here,$A = (c^2 - ab)$,$B = -2(a^2 - bc)$,and $C = (b^2 - ac)$.Setting $D = 0$:$[-2(a^2 - bc)]^2 - 4(c^2 - ab)(b^2 - ac) = 0$$4(a^2 - bc)^2 - 4(c^2 - ab)(b^2 - ac) = 0$Dividing by $4$:$(a^4 - 2a^2bc + b^2c^2) - (b^2c^2 - ac^3 - ab^3 + a^2bc) = 0$$a^4 - 2a^2bc + b^2c^2 - b^2c^2 + ac^3 + ab^3 - a^2bc = 0$$a^4 - 3a^2bc + ac^3 + ab^3 = 0$Factoring out $a$:$a(a^3 + b^3 + c^3 - 3abc) = 0$Thus,the condition is $a = 0$ or $a^3 + b^3 + c^3 = 3abc$.

The condition for the roots of the equation $(c^2 - ab)x^2 - 2(a^2 - bc)x + (b^2 - ac) = 0$ to be equal is

Similar Questions

If ${x^2} + 2x + 2xy + my - 3$ has two rational factors,then the value of $m$ will be

If $\alpha$ and $\beta$ are roots of the equation $Ax^2 + Bx + C = 0$,then the value of $\alpha^3 + \beta^3$ is

If $x^{4}+\frac{1}{x^{4}}=119$ and $x>1$,then find the positive value of $x^{3}-\frac{1}{x^{3}}$.

The sum of the roots of the equation,${x^2} + |2x - 3| - 4 = 0$,is

Solve the given two equations and select the correct option.
$I.$ $7x - 3y = 13$
$II.$ $5x + 4y = 40$

Vedclass Test Series

Exam Paper Generator

Online Exam Module