Question

If a root of the equation $a{x}^2+bx+c=0$ be reciprocal of the equation then $a^{\prime }{x}^2+b^{\prime }x+c^{\prime }=0$, then

• A. $(c{c}^{\prime }-a{a}^{\prime }{)}^2=(b{a}^{\prime }-c{b}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$
• B. $(b{b}^{\prime }-a{a}^{\prime }{)}^2=(c{a}^{\prime }-b{c}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$
• C. $(c{c}^{\prime }-a{a}^{\prime }{)}^2=(b{a}^{\prime }-c{b}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$
• D. $(c{c}^{\prime }+a{a}^{\prime }{)}^2=(b{a}^{\prime }-c{b}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$

Answer 1

The correct option is A

$(c{c}^{\prime }-a{a}^{\prime }{)}^2=(b{a}^{\prime }-c{b}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$

Let α be a root of first equation, then $\frac{1}{\alpha }$ be a root of second equation.

therefore $a{\alpha }^2+b\alpha +c=0$ and $a^{\prime }\frac{1}{{\alpha }^2}+b^{\prime }\frac{1}{\alpha }+c^{\prime }=0$ or $c^{\prime }{\alpha }^2+b^{\prime }\alpha +d=0$

Hence $\frac{{\alpha }^2}{b{a}^{\prime }-b^{\prime }c}=\frac{\alpha }{c{c}^{\prime }-a{a}^{\prime }}=\frac{1}{a{b}^{\prime }-b{c}^{\prime }}$

$(c{c}^{\prime }-a{a}^{\prime }{)}^2=(b{a}^{\prime }-c{b}^{\prime }\left)\right(a{b}^{\prime }-b{c}^{\prime }).$