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Question Number 50630 by peter frank last updated on 18/Dec/18

Answered by ajfour last updated on 18/Dec/18

$$\alpha +\beta =0,\gamma +\delta =-b...\left(i\right)$$$$\alpha \beta +(\alpha +\beta )\gamma +\gamma \delta +(\alpha +\beta )\delta =c$$$$\Rightarrow \alpha \beta +\gamma \delta =c....\left(ii\right)$$$$\alpha \beta .\gamma \delta =e....\left(iii\right)$$$$e(\frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }+\frac{1}{\delta })=-d$$$$\Rightarrow \frac{\gamma +\delta }{\gamma \delta }=\frac{-d}{e}....\left(iv\right)$$$$\Rightarrow \mathit{\gamma }\mathit{\delta }=\frac{be}{d},\mathit{\alpha }\mathit{\beta }=\frac{e}{\gamma \delta }=\frac{d}{b}$$$$usingin\left(ii\right)$$$$\frac{d}{b}+\frac{be}{d}=c$$$$\Rightarrow {\mathit{b}}^2\mathit{e}+{\mathit{d}}^2=\mathit{b}\mathit{c}\mathit{d}.$$

Commented by peter frank last updated on 20/Dec/18

$$thankyousir...pleasehelp$$$$50754$$

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