Question

Given that $\alpha$, $\gamma$ are the roots of equations $A{x}^2$ - 4x + 1 = 0 and $\beta$, $\gamma$ the roots of equation $B{x}^2$ - 6x + 1 = 0. then find the value of (A +B), such that $\alpha$ , $\beta$ , $\gamma$ & $\delta$ are in H.P.

Answer 1

$A{x}^2-4x+1=0\alpha +\gamma =\frac{4}{A}\alpha \gamma =\frac{1}{A}B{x}^2-6x+1=0\beta +\gamma =\frac{6}{B}\beta \gamma =\frac{1}{B}Also\alpha ,\beta ,\gamma ,\delta areinH.P.⟹\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma },\frac{1}{\delta }areinA.P.then,wehave:\frac{1}{\beta }=(\frac{1}{\alpha }+\frac{1}{\gamma })\frac{1}{2}⟹\frac{2}{\beta }=\left(\frac{\alpha +\gamma }{\alpha \gamma }\right)⟹\beta =\frac{2\alpha \gamma }{\alpha +\gamma }⟹\beta =2\times \frac{1}{A}\times \frac{A}{4}=\frac{1}{2}\beta \gamma =\frac{1}{B}⟹\gamma =\frac{2}{B}\beta +\gamma =\frac{6}{B}⟹\frac{1}{2}+\frac{2}{B}=\frac{6}{B}⟹\frac{1}{2}=\frac{6}{B}-\frac{2}{B}=\frac{4}{B}⟹B=4\times 2=8\gamma =\frac{2}{B}=\frac{2}{8}=\frac{1}{4}\alpha \gamma =\frac{1}{A}⟹\alpha \cdot \frac{1}{4}=\frac{1}{A}⟹\alpha =\frac{4}{A}\alpha +\gamma =\frac{4}{A}⟹\gamma =0$

Also according to A.P., we have,

$\frac{1}{\beta }-\frac{1}{\alpha }=\frac{1}{\gamma }-\frac{1}{\beta }\alpha +\gamma =\frac{4}{A}⟹\alpha +\frac{2}{B}=\frac{4}{A}⟹\alpha =\frac{4}{A}-\frac{2}{B}=\frac{4}{A}-\frac{2}{8}⟹\alpha =\frac{4}{A}-\frac{2}{8}=\frac{32-2A}{8A}\alpha \gamma =\frac{1}{A}⟹\alpha \cdot \frac{2}{B}=\frac{1}{A}⟹\alpha \cdot \frac{2}{4}=\frac{1}{A}⟹\alpha =\frac{2}{A}\therefore \frac{2}{A}=\frac{32-2A}{8A}⟹16=32-2A⟹2A=32-16=16⟹A=\frac{16}{2}=8\therefore (A+B)=8+8=16$