If α and β are the roots of the equation, 7x^2-3x-2=0, then the value ofα 1-α ^2+ β 1-β ^2 is equal to:

Given $\alpha$α and $\beta$β are the roots of the equation,$7{x}^{2}-3x-2=0,$7x2−3x−2=0, then $\alpha +\beta =\frac{3}{7};\alpha \beta =-\frac{2}{7}$α+β=73​;αβ=−72​

Now

$$\frac{\alpha }{1-{\alpha }^{2}}+\frac{\beta }{1-{\beta }^{2}}=\frac{(\alpha +\beta )-\alpha \beta (\alpha +\beta )}{(1-{\alpha }^{2})(1-{\beta }^{2})}=\frac{(\alpha +\beta )-\alpha \beta (\alpha +\beta )}{1+(\alpha \beta {)}^{2}-({\alpha }^{2}+{\beta }^{2})}$$1−α2α​+1−β2β​=(1−α2)(1−β2)(α+β)−αβ(α+β)​=1+(αβ)2−(α2+β2)(α+β)−αβ(α+β)​

$$\Rightarrow \frac{(\alpha +\beta )-\alpha \beta (\alpha +\beta )}{1+(\alpha \beta {)}^{2}-(\alpha +\beta {)}^{2}+2\alpha \beta }=\frac{\frac{3}{7}+\frac{2}{7}(\frac{3}{7})}{1+{(\frac{2}{7})}^{2}-{(\frac{3}{7})}^{2}-2(\frac{2}{7})}=\frac{27}{16}$$⇒1+(αβ)2−(α+β)2+2αβ(α+β)−αβ(α+β)​=1+(72​)2−(73​)2−2(72​)73​+72​(73​)​=1627​