Question

If $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0,{\alpha }_1,-\beta$ are the roots of $a_1{x}^2+b_{1}x+c_1=0$, show that $\alpha ,{\alpha }_1$ are the roots of $\frac{x^2}{\frac{b}{a}+\frac{b_1}{a_1}}+x+\frac{1}{\frac{b}{c}+\frac{b_1}{c_1}}=0$.

Answer 1

$\alpha +\beta =-b/a$.........(1)

$\alpha \beta =c/a$............(2)

${\alpha }_1-\beta =-b_1/a_1$..........(3)

$-{\alpha }_1\beta =c_1/a_1$..........(4)

Adding (1) and (3),

$\alpha +{\alpha }_1=-\frac{b}{a}+\frac{b_1}{a_1}=S$. ............(5)

Dividing (1) by (2) and (3) by (4), we get

$\frac{1}{\beta }+\frac{1}{\alpha }=-\frac{b}{c}$ and $-\frac{1}{\beta }+\frac{1}{{\alpha }_1}=-\frac{b_1}{c_1}$

Adding them, we get

$\frac{1}{\alpha }+\frac{1}{{\alpha }_1}=-\frac{b}{c}-\frac{b_1}{c_1}$

or $\frac{\alpha +{\alpha }_1}{\alpha {\alpha }_1}=\frac{S}{P}=-\frac{b}{c}-\frac{b_1}{c_1}$.........(6)

$\therefore$ The equation is $x^2-xS+P=0$

or $\frac{x^2}{-S}+x-\frac{P}{S}=0$. Now use (5) and (6)