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Question Number 55658 by otchereabdullai@gmail.com last updated on 01/Mar/19

$$\mathrm{find}\mathrm{the}\mathrm{difference}\mathrm{of}\mathrm{the}\mathrm{roots}\mathrm{of}\mathrm{the}$$$$\mathrm{following}\mathrm{quadratic}\mathrm{equation}$$$$(3+2\sqrt{2}){\mathrm{x}}^2+(1+\sqrt{2})\mathrm{x}=2$$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19

$$\alpha +\beta =-\frac{(1+\sqrt{2})}{3+2\sqrt{2}}\alpha \beta =\frac{-2}{3+2\sqrt{2}}$$$${(\alpha -\beta )}^2={(\alpha +\beta )}^2-4\alpha \beta$$$$=\frac{1+2\sqrt{2}+2}{{(3+2\sqrt{2})}^2}-4\left(\frac{-2}{3+2\sqrt{2}}\right)$$$$=\frac{3+2\sqrt{2}}{{(3+2\sqrt{2})}^2}+\frac{8}{3+2\sqrt{2}}$$$$=\frac{1}{3+2\sqrt{2}}+\frac{8}{3+2\sqrt{2}}$$$$=\frac{9}{3+2\sqrt{2}}$$$$(\alpha -\beta )=\sqrt{\frac{9}{3+2\sqrt{2}}}$$$$=\frac{3}{\sqrt{2}+1}$$

Commented by otchereabdullai@gmail.com last updated on 01/Mar/19

$$\mathrm{Thank}\mathrm{you}\mathrm{Prof}$$

Commented by mr W last updated on 01/Mar/19

$$pleasechecksir,shoulditnotbe:$$$$\frac{3}{\sqrt{2}+1}=3(\sqrt{2}-1)?$$

Commented by tanmay.chaudhury50@gmail.com last updated on 01/Mar/19

$$yessir..inhurry...$$

Commented by malwaan last updated on 02/Mar/19

$$\sqrt{\frac{9(3-2\sqrt{2})}{9-8}}=3\sqrt{3-2\sqrt{2}}$$

Commented by mr W last updated on 02/Mar/19

$$and3\sqrt{3-2\sqrt{2}}=3\sqrt{{(\sqrt{2}-1)}^2}=3(\sqrt{2}-1)$$

Commented by otchereabdullai@gmail.com last updated on 02/Mar/19

$$\mathrm{thanks}\mathrm{Sir}\mathrm{malwaan}\mathrm{and}\mathrm{Prof}\mathrm{W}\mathrm{for}$$$$\mathrm{the}\mathrm{correction}$$

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