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Question Number 35512 by Rio Mike last updated on 19/May/18

$$Giventhat(x+1,3,x)arelengths$$$$ofthesidesofarightangled$$$$triangle\left(pythagorastripple\right)$$$$findthevalueofx.$$

Answered by Rasheed.Sindhi last updated on 20/May/18

$$\mathrm{Case}-1:\mathrm{x}+1>3$$$$\therefore \mathrm{x}+1\mathrm{is}\mathrm{hypatenuse}$$$${(\mathrm{x}+1)}^2={\left(3\right)}^2+{\mathrm{x}}^2$$$${\mathrm{x}}^2+2\mathrm{x}+1=9+{\mathrm{x}}^2$$$$2\mathrm{x}=9-1=8$$$$\mathrm{x}=4$$$$\mathrm{x}+1=4+1=5$$$$(5,3,4)\mathrm{is}\mathrm{required}\mathrm{pythagorean}\mathrm{triplet}.$$$$\mathrm{Case}-2:3>\mathrm{x}+1$$$$\therefore 3\mathrm{is}\mathrm{hypatenuse}$$$${\left(3\right)}^2={(\mathrm{x}+1)}^2+{\mathrm{x}}^2$$$${2\mathrm{x}}^2+2\mathrm{x}-8=0$$$${\mathrm{x}}^2+\mathrm{x}-4=0$$$$\mathrm{x}=\frac{-1\pm \sqrt{17}}{2}$$$$\mathrm{x}+1=\frac{-1\pm \sqrt{17}}{2}+1=\frac{1\pm \sqrt{17}}{2}$$$$(\frac{1+\sqrt{17}}{2},3,\frac{-1+\sqrt{17}}{2})\mathrm{contains}\mathrm{irrational}\mathrm{numbers}$$$$(\frac{1-\sqrt{17}}{2},3,\frac{-1-\sqrt{17}}{2})\mathrm{contains}\mathrm{negative}$$$$\mathrm{numbers}.$$$$\mathrm{Hence}$$$$(5,3,4)\mathrm{is}\mathrm{required}\mathrm{pythagorean}\mathrm{triplet}.$$$$$$

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