Question

If the equation $\frac{1}{x}+\frac{1}{x+\alpha }=\frac{1}{\lambda }+\frac{1}{\lambda +\alpha }$ has real roots that are equal in magnitude and opposite in sign, then

• A. ${\lambda }^2=3a^2$
• B. ${\lambda }^2=2a^2$
• C. ${\lambda }^2=a^2$
• D. $a^2=2{\lambda }^2$

Answer 1

The correct option is D $a^2=2{\lambda }^2$

$\frac{1}{x}+\frac{1}{x+a}=\frac{1}{\lambda }+\frac{1}{\lambda +\alpha }\frac{x+a+x}{x^2+ax}=\frac{1}{\lambda }+\frac{1}{\lambda +\alpha }\frac{2x+a}{x^2+ax}=\frac{1}{\lambda }+\frac{1}{\lambda +\alpha }\frac{2x+a}{x^2+ax}=\frac{2\lambda +a}{{\lambda }^2+a\lambda }({\lambda }^2+a\lambda )(2x+a)=(x^2+ax)(2\lambda +a)(2\lambda {+a)x}^2+(2\lambda a+a^2-2{\lambda }^2-2a\lambda )x-a{\lambda }^2-a^2\lambda =0(2\lambda {+a)x}^2+(a^2-2{\lambda }^2)x-a{\lambda }^2-a^2\lambda =0$

Now roots are equal in magnitude and opposite in sign so their sum is zero

$-\frac{a^2-2{\lambda }^2}{2\lambda +a}=0a^2=2{\lambda }^2$