Question

The condition that the equation 1/x + 1/x+b = 1/m + 1/m+b has real roots that are equal in magnitude but opposite in sign

Answer 1

Answer:

The required condition is

${b}^{2} = 2 {m}^{2}$

Step-by-step explanation:

$\frac{1}{x} + \frac{1}{x + b} = \frac{1}{m} + \frac{1}{m + b}$

$\frac{(x + b) + x}{x( x+ b)} = \frac{(m + b) + m}{m(m+ b)}$

$\frac{2x + b}{x(x + b)} = \frac{2m + b}{m(m + b)}$

$(2x + b)m(m + b) = (2m + b)x(x + b)$

$2m(m + b)x + mb(m + b) = (2m + b) {x}^{2} + (2m + b)bx$

$(2m + b) {x}^{2} + ((2m + b)b - 2m(m + b))x - mb(m + b) = 0$

It is given that roots are equal and opposite ,

$or \: \alpha = - \beta$

$or \: \alpha + \beta = 0$

$or \: - \frac{coeff. \: of \: x}{coeff. \: of \: {x}^{2} } = 0$

$\frac{(2m + b)b - 2m(m + b)}{2m + b} = 0$

$2mb + {b}^{2} - 2 {m}^{2} - 2mb = 0$

${b}^{2} - 2 {m}^{2} = 0$

${b}^{2} = 2 {m}^{2}$

This is the required condition.