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Question Number 68601 by Maclaurin Stickker last updated on 14/Sep/19

ABCD is a side square 1. B, F and E are collinear. FDE is a right triangle with hypotenuse 1 and the DE cathetus is worth x. What is the value of x? (Solve with algebra)

$${ABCD}\:\mathrm{is}\:\mathrm{a}\:\mathrm{side}\:\mathrm{square}\:\mathrm{1}.\: \\ $$$${B},\:{F}\:\mathrm{and}\:{E}\:\mathrm{are}\:\mathrm{collinear}. \\ $$$${FDE}\:\mathrm{is}\:\mathrm{a}\:\mathrm{right}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{hypotenuse}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{the}\:{DE}\:\mathrm{cathetus}\:\mathrm{is}\:\mathrm{worth}\:\boldsymbol{{x}}.\: \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\boldsymbol{{x}}? \\ $$$$\left(\mathrm{Solve}\:\mathrm{with}\:\mathrm{algebra}\right) \\ $$

Commented by Maclaurin Stickker last updated on 14/Sep/19

Commented by mr W last updated on 14/Sep/19

BE=(√((1+x)^2 +1^2 ))=(√(x^2 +2x+2)) (1/x)=((√(x^2 +2x+2))/(1+x)) ((1+2x+x^2 )/x^2 )=x^2 +2x+2 x^4 +2x^3 +2x^2 =1+2x+x^2 x^4 +2x^3 +x^2 −2x−1=0 ⇒x≈0.8832 (exact value possible, but complicated)

$${BE}=\sqrt{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} +\mathrm{1}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{{x}}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2}}}{\mathrm{1}+{x}} \\ $$$$\frac{\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{2} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}\approx\mathrm{0}.\mathrm{8832} \\ $$$$\left({exact}\:{value}\:{possible},\:{but}\:{complicated}\right) \\ $$

Commented by MJS last updated on 14/Sep/19

exact value, just for the fun of it x^4 +2x^3 +x^2 −2x−1=0 (x^2 +(1−(√2))x+1−(√2))(x^2 +(1+(√2))x+1+(√2))=0 x_1 =−(1/2)+((√2)/2)−((√(−1+2(√2)))/2)<0 ⇒ not valid x_2 =−(1/2)+((√2)/2)+((√(−1+2(√2)))/2)>0 ⇒ is the answer x_(3, 4) =−(1/2)−((√2)/2)±((√(1+2(√2)))/2)i ∉R

$$\mathrm{exact}\:\mathrm{value},\:\mathrm{just}\:\mathrm{for}\:\mathrm{the}\:\mathrm{fun}\:\mathrm{of}\:\mathrm{it} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}+\mathrm{1}−\sqrt{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){x}+\mathrm{1}+\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}<\mathrm{0}\:\Rightarrow\:\mathrm{not}\:\mathrm{valid} \\ $$$${x}_{\mathrm{2}} =−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}>\mathrm{0}\:\Rightarrow\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer} \\ $$$${x}_{\mathrm{3},\:\mathrm{4}} =−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\mathrm{i}\:\notin\mathbb{R} \\ $$

Commented by Maclaurin Stickker last updated on 14/Sep/19

Ha, ha! Thank you, sir!

$$\mathrm{Ha},\:\mathrm{ha}!\:\mathrm{Thank}\:\mathrm{you},\:\mathrm{sir}! \\ $$

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