If a root of the given equation a(b - c)x^2 + | vedclass

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

Question

(C) Given the quadratic equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Let the roots be $\alpha$ and $\beta$. We are given $\alpha = 1$.The product

Vedclass · Accepted Answer

$\frac{c(a - b)}{a(b - c)}$

Vedclass · Answer

(C) Given the quadratic equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$.Let the roots be $\alpha$ and $\beta$. We are given $\alpha = 1$.The product of the roots of a quadratic equation $Ax^2 + Bx + C = 0$ is given by $\frac{C}{A}$.Here,$A = a(b - c)$,$B = b(c - a)$,and $C = c(a - b)$.Therefore,$\alpha \cdot \beta = \frac{c(a - b)}{a(b - c)}$.Since $\alpha = 1$,we have $1 \cdot \beta = \frac{c(a - b)}{a(b - c)}$.Thus,$\beta = \frac{c(a - b)}{a(b - c)}$.

If a root of the given equation $a(b - c)x^2 + b(c - a)x + c(a - b) = 0$ is $1$,then the other root will be

Similar Questions

Form a quadratic equation whose one root is $3-\sqrt{5}$ and the sum of roots is $6$.

If $\alpha$ and $\beta$ are two real numbers satisfying $\alpha^2 + \beta^2 = 5$ and $3(\alpha^5 + \beta^5) = 11(\alpha^3 + \beta^3)$,then the value of $\alpha \beta$ is:

If the equation $\frac{x^2 + 5}{2} = x - 2\cos(ax + b)$ has at least one solution,then $(b + a)$ can be equal to

Let $S$ be the set of all real roots of the equation,$3^{x}(3^{x}-1)+2=|3^{x}-1|+|3^{x}-2|$. Then $S$

The value of $p$ for which the sum of the squares of the roots of the equation $x^2 - (p + 3)x + (5p - 2) = 0$ assumes its least value is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module