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Question Number 205502 by MATHEMATICSAM last updated on 22/Mar/24

$$\mathrm{If}\mathrm{two}\mathrm{roots}\mathrm{of}{ax}^2+bx+c=0\mathrm{are}\alpha \mathrm{and}$$$$\beta \mathrm{then}\frac{1}{{(a{\alpha }^2+c)}^2}+\frac{1}{{(a{\beta }^2+c)}^2}=?$$

Answered by Rasheed.Sindhi last updated on 23/Mar/24

$$\mathrm{Another}\mathrm{Way}$$$$\frac{1}{{(a{\alpha }^2+c)}^2}+\frac{1}{{(a{\beta }^2+c)}^2}$$$$=\frac{1}{a^2{({\alpha }^2+\frac{c}{a})}^2}+\frac{1}{a^2{({\beta }^2+\frac{c}{a})}^2}$$$$=\frac{1}{a^2{({\alpha }^2+\alpha \beta )}^2}+\frac{1}{a^2{({\beta }^2+\alpha \beta )}^2}$$$$=\frac{1}{a^2{\alpha }^2{(\alpha +\beta )}^2}+\frac{1}{a^2{\beta }^2{(\beta +\alpha )}^2}$$$$=\frac{1}{a^2{(\alpha +\beta )}^2}(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2})$$$$=\frac{1}{a^2{(\alpha +\beta )}^2}\cdot \frac{{\alpha }^2+{\beta }^2}{{\left(\alpha \beta \right)}^2}$$$$=\frac{1}{a^2{(\alpha +\beta )}^2}\cdot \frac{{(\alpha +\beta )}^2-2\alpha \beta }{{\left(\alpha \beta \right)}^2}$$$$=\frac{1}{a^2{(-\frac{b}{a})}^2}\cdot \frac{{(-\frac{b}{a})}^2-2\left(\frac{c}{a}\right)}{{\left(\frac{c}{a}\right)}^2}$$$$=\frac{1}{b^2}\cdot \frac{\frac{b^2}{a^2}-\frac{2c}{a}}{\frac{c^2}{a^2}}$$$$=\frac{1}{b^2}\cdot \frac{\frac{b^2-2ac}{a^2}}{\frac{c^2}{a^2}}$$$$=\frac{b^2-2ac}{b^2{c}^2}$$

Answered by A5T last updated on 22/Mar/24

$$a{\alpha }^2+b\alpha +c=0;a{\beta }^2+b\beta +c=0$$$$\Rightarrow ?=\frac{1}{{(-b\alpha )}^2}+\frac{1}{{(-b\beta )}^2}=\frac{1}{b^2}\left(\frac{{\alpha }^2+{\beta }^2}{{\left(\alpha \beta \right)}^2}\right)=\frac{\frac{b^2}{a^2}-\frac{2c}{a}}{b^2\left(\frac{c^2}{a^2}\right)}$$$$=\frac{b^2-2ac}{b^2{c}^2}$$

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