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Question Number 69385 by Maclaurin Stickker last updated on 23/Sep/19
Answered by Maclaurin Stickker last updated on 23/Sep/19
There is another way to solve this question just using algebra: ((FD)/1)=(x/(x+1))⇒FD=(x/(x+1)) Now we can use Pythagorean theorem on △FDE: FD^2 +DE^2 =FE^2 ⇒(x^2 /((x+1)^2 ))+x^2 =1 ⇒Now we can multiply the sides by (x+1)^2 : x^2 +x^2 (x+1)^2 =(x+1)^2 ⇒(x(x+1))^2 =(x+1)^2 −x^2 Now we can use the formula (a+b)^2 =a^2 +2ab+b^2 It is square completion. let b=1 and a=x(x+1) (x(x+1))^2 +2x(x+1)1+1^2 =(x+1)^2 −x^2 +2x(x+1)1+1^2 We can convert the first expression to a perfect square trinomial (x^2 +x+1)^2 =x^2 +2x+1−x^2 +2x(x+1)+1 ⇒(x^2 +x+1)^2 =2x^2 +4x+2 ⇒(x^2 +x+1)^2 =2(x^2 +2x+1) (x^2 +x+1)^2 =2(x+1)^2 ⇒x^2 +x+1=(√(2(x+1)^2 )) ⇒x^2 +x+1=(x+1)(√(2 )) ⇒x^2 +x+1=x(√2)+(√2) ⇒x^2 +x+1−x(√2)−(√2)=0 ⇒x^2 +(1−(√2))x+1−(√2)=0 Using quadratic formula, we have: x=((−(1−(√2))±(√((1−(√2))^2 −4.1.(1−(√2)))))/2) ⇒x=((−1+(√2)±(√(−1−2(√2)+4(√2))))/2) ⇒x_1 =((−1+(√2)+(√(−1+2(√2))))/2) and x_2 =((−1+(√2)−(√(−1+2(√2))))/2) x_(2 ) is negative, then the answer is x_1 =((−1+(√2)+(√(−1+2(√2))))/2)≈0.883204
$${There}\:{is}\:{another}\:{way}\:{to}\:{solve}\:{this} \\ $$$${question}\:{just}\:{using}\:{algebra}: \\ $$$$\frac{{FD}}{\mathrm{1}}=\frac{{x}}{{x}+\mathrm{1}}\Rightarrow{FD}=\frac{{x}}{{x}+\mathrm{1}}\:{Now}\:{we}\:{can}\:{use} \\ $$$${Pythagorean}\:{theorem}\:{on}\:\bigtriangleup{FDE}: \\ $$$${FD}^{\mathrm{2}} +{DE}^{\mathrm{2}} ={FE}^{\mathrm{2}} \Rightarrow\frac{{x}^{\mathrm{2}} }{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }+{x}^{\mathrm{2}} =\mathrm{1} \\ $$$$\Rightarrow{Now}\:{we}\:{can}\:{multiply}\:{the}\:{sides}\:{by}\:\left({x}+\mathrm{1}\right)^{\mathrm{2}} : \\ $$$${x}^{\mathrm{2}} +{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\left({x}\left({x}+\mathrm{1}\right)\right)^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} \\ $$$${Now}\:{we}\:{can}\:{use}\:{the}\:{formula}\:\left({a}+{b}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +\mathrm{2}{ab}+{b}^{\mathrm{2}} \\ $$$${It}\:{is}\:{square}\:{completion}. \\ $$$${let}\:{b}=\mathrm{1}\:{and}\:{a}={x}\left({x}+\mathrm{1}\right) \\ $$$$\left({x}\left({x}+\mathrm{1}\right)\right)^{\mathrm{2}} +\mathrm{2}{x}\left({x}+\mathrm{1}\right)\mathrm{1}+\mathrm{1}^{\mathrm{2}} =\left({x}+\mathrm{1}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} +\mathrm{2}{x}\left({x}+\mathrm{1}\right)\mathrm{1}+\mathrm{1}^{\mathrm{2}} \\ $$$${We}\:{can}\:{convert}\:{the}\:{first}\:{expression} \\ $$$${to}\:{a}\:{perfect}\:{square}\:{trinomial} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{2}{x}\left({x}+\mathrm{1}\right)+\mathrm{1} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2} \\ $$$$\Rightarrow\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right) \\ $$$$\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} \Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}=\sqrt{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}=\left({x}+\mathrm{1}\right)\sqrt{\mathrm{2}\:} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}={x}\sqrt{\mathrm{2}}+\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{x}+\mathrm{1}−{x}\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){x}+\mathrm{1}−\sqrt{\mathrm{2}}=\mathrm{0} \\ $$$${Using}\:{quadratic}\:{formula},\:{we}\:{have}: \\ $$$${x}=\frac{−\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\pm\sqrt{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{4}.\mathrm{1}.\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{2}}\pm\sqrt{−\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\:{and}\:{x}_{\mathrm{2}} =\frac{−\mathrm{1}+\sqrt{\mathrm{2}}−\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}\:} \:{is}\:{negative},\:{then}\:{the}\:{answer}\:{is} \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}}\approx\mathrm{0}.\mathrm{883204} \\ $$$$ \\ $$
Answered by MJS last updated on 23/Sep/19
coordinate method let x=q to not get confused with x, y coordinates A= ((0),(0) ) B= ((0),(1) ) C= ((1),(1) ) D= ((1),(0) ) E= (((1+q)),(0) ) line BE: y=−(1/(q+1))x+1 F∈BE ⇒ F= ((1),((q/(q+1))) ) ∣EF∣^2 =1 q^2 +(q^2 /((q+1)^2 ))=1 q^4 +2q^3 +q^2 −2q−1=0 q=t−(1/2) t^4 −(1/2)t^2 −2t+(1/(16))=0 (t^2 −(√2)t+((3−2(√2))/4))(t^2 +(√2)t+((3+2(√2))/4))=0 ⇒ t=((√2)/2)±((√(−1+2(√2)))/2) x>0 ⇒ x=−((1+(√2)+(√(−1+2(√2))))/2)
$$\mathrm{coordinate}\:\mathrm{method} \\ $$$$\mathrm{let}\:{x}={q}\:\mathrm{to}\:\mathrm{not}\:\mathrm{get}\:\mathrm{confused}\:\mathrm{with}\:{x},\:{y}\:\mathrm{coordinates} \\ $$$${A}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:\:{B}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\:\:{C}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\end{pmatrix}\:\:{D}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix} \\ $$$${E}=\begin{pmatrix}{\mathrm{1}+{q}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\mathrm{line}\:{BE}:\:{y}=−\frac{\mathrm{1}}{{q}+\mathrm{1}}{x}+\mathrm{1} \\ $$$${F}\in{BE}\:\Rightarrow\:{F}=\begin{pmatrix}{\mathrm{1}}\\{\frac{{q}}{{q}+\mathrm{1}}}\end{pmatrix} \\ $$$$\mid{EF}\mid^{\mathrm{2}} =\mathrm{1} \\ $$$${q}^{\mathrm{2}} +\frac{{q}^{\mathrm{2}} }{\left({q}+\mathrm{1}\right)^{\mathrm{2}} }=\mathrm{1} \\ $$$${q}^{\mathrm{4}} +\mathrm{2}{q}^{\mathrm{3}} +{q}^{\mathrm{2}} −\mathrm{2}{q}−\mathrm{1}=\mathrm{0} \\ $$$${q}={t}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}^{\mathrm{4}} −\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} −\mathrm{2}{t}+\frac{\mathrm{1}}{\mathrm{16}}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\sqrt{\mathrm{2}}{t}+\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{2}}{t}+\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{4}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{t}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$$${x}>\mathrm{0} \\ $$$$\Rightarrow\:{x}=−\frac{\mathrm{1}+\sqrt{\mathrm{2}}+\sqrt{−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}}}}{\mathrm{2}} \\ $$
Commented by Mr. K last updated on 23/Sep/19
Great!
$${Great}! \\ $$
Commented by MJS last updated on 23/Sep/19
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$

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