For the equation frac{1}{x + a} - frac{1}{x + | vedclass

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

Question

(D) Given the equation: $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$Simplify the left side: $\frac{(x + b) - (x + a)}{(x + a)(x + b)} = \frac

Vedclass · Accepted Answer

$\frac{-2bc}{b + c}$

Vedclass · Answer

(D) Given the equation: $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$Simplify the left side: $\frac{(x + b) - (x + a)}{(x + a)(x + b)} = \frac{1}{x + c}$$\frac{b - a}{x^2 + (a + b)x + ab} = \frac{1}{x + c}$Cross-multiply: $x^2 + (a + b)x + ab = (b - a)(x + c)$$x^2 + (a + b)x + ab = (b - a)x + (b - a)c$$x^2 + (a + b - b + a)x + ab - (b - a)c = 0$$x^2 + 2ax + ab - bc + ac = 0$Since the product of the roots is $0$,the constant term must be $0$:$ab + ac - bc = 0 \implies a(b + c) = bc \implies a = \frac{bc}{b + c}$The sum of the roots of a quadratic equation $x^2 + Bx + C = 0$ is $-B$.Sum of roots $= -2a = -2 \left( \frac{bc}{b + c} \right) = \frac{-2bc}{b + c}$.

For the equation $\frac{1}{x + a} - \frac{1}{x + b} = \frac{1}{x + c}$,if the product of the roots is zero,what is the sum of the roots?

Similar Questions

The cubic equation whose roots are the squares of the roots of $x^3-2x^2+10x-8=0$ is

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

If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $x^{2}-bx+c=0$, then which of the following statements is/are correct?

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3+ax^2+bx+c=0$,then $\alpha^{-1}+\beta^{-1}+\gamma^{-1}$ is equal to

The difference between two roots of the equation $x^3-13x^2+15x+189=0$ is $2$. Then the roots of the equation are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module