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

Given that (x+1,3,x) are lengths of the sides of a right angled triangle(pythagoras tripple) find the value of x.

$$\:{Given}\:{that}\:\left({x}+\mathrm{1},\mathrm{3},{x}\right)\:{are}\:{lengths} \\ $$$${of}\:{the}\:{sides}\:{of}\:{a}\:{right}\:{angled} \\ $$$${triangle}\left({pythagoras}\:{tripple}\right) \\ $$$${find}\:{the}\:{value}\:{of}\:{x}. \\ $$

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

Case-1: x+1>3 ∴ x+1 is hypatenuse (x+1)^2 =(3)^2 +x^2 x^2 +2x+1=9+x^2 2x=9−1=8 x=4 x+1=4+1=5 (5,3,4) is required pythagorean triplet. Case-2:3>x+1 ∴ 3 is hypatenuse (3)^2 =(x+1)^2 +x^2 2x^2 +2x−8=0 x^2 +x−4=0 x=((−1±(√(17)))/2) x+1=((−1±(√(17)))/2)+1=((1±(√(17)))/2) (((1+(√(17)))/2) ,3,((−1+(√(17)))/2)) contains irrational numbers (((1−(√(17)))/2) ,3,((−1−(√(17)))/2)) contains negative numbers. Hence (5,3,4) is required pythagorean triplet.

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

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