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Question Number 36221 Answers: 2 Comments: 1

a(b+(1/b))=−1 ....(i) b−(1/b)=a^2 ....(ii) Solve simultaneously for a, and b.

$${a}\left({b}+\frac{\mathrm{1}}{{b}}\right)=−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({i}\right) \\ $$$${b}−\frac{\mathrm{1}}{{b}}={a}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\left({ii}\right) \\ $$$${Solve}\:{simultaneously}\:{for}\:{a},\:{and}\:{b}. \\ $$

Question Number 36219 Answers: 0 Comments: 2

in question 34992 this had to be solved: t^4 +8t^3 +2t^2 −8t+1=0 here another concept worked (I tried some) t_1 =a+b+c+d t_2 =a+b−c−d t_3 =a−b+c−d t_3 =a−b−c+d (t−t_1 )(t−t_2 )(t−t_3 )(t−t_4 )=0 leads to t^4 −4at^3 +2(3a^2 −b^2 −c^2 −d^2 )t^2 + +4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)t+ +a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 ) 1. −4a=8 2. 2(3a^2 −b^2 −c^2 −d^2 )=2 3. 4(a(−a^2 +b^2 +c^2 +d^2 )−2bcd)=−8 4. a^4 +b^4 +c^4 +d^4 −2(a^2 (b^2 +c^2 +d^2 )+b^2 (c^2 +d^2 )+c^2 d^2 )=1 1. a=−2 2. b^2 +c^2 +d^2 =11 3. b^2 +c^2 +d^2 +bcd=5 4. ... 3. −2. bcd=−6 now I tried by chance if there are easy real solutions b^2 =p, c^2 =q, d^2 =r, 0≤p≤q≤r p+q+r=11 this leads to p=2, q=3, p=6 ∣b∣=(√2), ∣c∣=(√3), ∣d∣=(√6) with one or three out of a, b, c negative because of bcd=−6 because of the nature of t_1 , t_2 , t_3 , t_4 these lead to the same solutions I took b=−(√2), c=−(√3), d=−(√6) t_1 =−2−(√2)−(√3)−(√6) t_2 =−2−(√2)+(√3)+(√6) t_3 =−2+(√2)−(√3)+(√6) t_4 =−2+(√2)+(√3)−(√6)

$$\mathrm{in}\:\mathrm{question}\:\mathrm{34992}\:\mathrm{this}\:\mathrm{had}\:\mathrm{to}\:\mathrm{be}\:\mathrm{solved}: \\ $$$${t}^{\mathrm{4}} +\mathrm{8}{t}^{\mathrm{3}} +\mathrm{2}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{here}\:\mathrm{another}\:\mathrm{concept}\:\mathrm{worked}\:\left(\mathrm{I}\:\mathrm{tried}\:\mathrm{some}\right) \\ $$$${t}_{\mathrm{1}} ={a}+{b}+{c}+{d} \\ $$$${t}_{\mathrm{2}} ={a}+{b}−{c}−{d} \\ $$$${t}_{\mathrm{3}} ={a}−{b}+{c}−{d} \\ $$$${t}_{\mathrm{3}} ={a}−{b}−{c}+{d} \\ $$$$\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)\left({t}−{t}_{\mathrm{3}} \right)\left({t}−{t}_{\mathrm{4}} \right)=\mathrm{0} \\ $$$$\mathrm{leads}\:\mathrm{to} \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{at}^{\mathrm{3}} +\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right){t}^{\mathrm{2}} + \\ $$$$\:\:\:\:\:+\mathrm{4}\left({a}\left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\mathrm{2}{bcd}\right){t}+ \\ $$$$\:\:\:\:\:+{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {d}^{\mathrm{2}} \right) \\ $$$$ \\ $$$$\mathrm{1}.\:−\mathrm{4}{a}=\mathrm{8} \\ $$$$\mathrm{2}.\:\mathrm{2}\left(\mathrm{3}{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$$\mathrm{3}.\:\mathrm{4}\left({a}\left(−{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)−\mathrm{2}{bcd}\right)=−\mathrm{8} \\ $$$$\mathrm{4}.\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} +{d}^{\mathrm{4}} −\mathrm{2}\left({a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} {d}^{\mathrm{2}} \right)=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=−\mathrm{2} \\ $$$$\mathrm{2}.\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\mathrm{11} \\ $$$$\mathrm{3}.\:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{bcd}=\mathrm{5} \\ $$$$\mathrm{4}.\:... \\ $$$$\mathrm{3}.\:−\mathrm{2}.\:{bcd}=−\mathrm{6} \\ $$$$ \\ $$$$\mathrm{now}\:\mathrm{I}\:\mathrm{tried}\:\mathrm{by}\:\mathrm{chance}\:\mathrm{if}\:\mathrm{there}\:\mathrm{are}\:\mathrm{easy}\:\mathrm{real}\:\mathrm{solutions} \\ $$$${b}^{\mathrm{2}} ={p},\:{c}^{\mathrm{2}} ={q},\:{d}^{\mathrm{2}} ={r},\:\mathrm{0}\leqslant{p}\leqslant{q}\leqslant{r} \\ $$$${p}+{q}+{r}=\mathrm{11} \\ $$$$\mathrm{this}\:\mathrm{leads}\:\mathrm{to} \\ $$$${p}=\mathrm{2},\:{q}=\mathrm{3},\:{p}=\mathrm{6} \\ $$$$\mid{b}\mid=\sqrt{\mathrm{2}},\:\mid{c}\mid=\sqrt{\mathrm{3}},\:\mid{d}\mid=\sqrt{\mathrm{6}}\:\mathrm{with}\:\mathrm{one}\:\mathrm{or}\:\mathrm{three}\:\mathrm{out} \\ $$$$\mathrm{of}\:{a},\:{b},\:{c}\:\mathrm{negative}\:\mathrm{because}\:\mathrm{of}\:{bcd}=−\mathrm{6} \\ $$$$\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:{t}_{\mathrm{1}} ,\:{t}_{\mathrm{2}} ,\:{t}_{\mathrm{3}} ,\:{t}_{\mathrm{4}} \:\mathrm{these} \\ $$$$\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{solutions} \\ $$$$\mathrm{I}\:\mathrm{took}\:{b}=−\sqrt{\mathrm{2}},\:{c}=−\sqrt{\mathrm{3}},\:{d}=−\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{1}} =−\mathrm{2}−\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{3}} =−\mathrm{2}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{6}} \\ $$$${t}_{\mathrm{4}} =−\mathrm{2}+\sqrt{\mathrm{2}}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{6}} \\ $$

Question Number 36207 Answers: 0 Comments: 4

x^4 +10x^3 +6x−1=0 for those who want an exact solution... (x−a−bi)(x−a+bi)(x−c−d)(x−c+d)=0 x^4 −2(a+c)x^3 +(a^2 +4ac+b^2 +c^2 −d^2 )x^2 − −2(a(c^2 −d^2 )+c(a^2 +b^2 ))x+(a^2 +b^2 )(c^2 −d^2 )=0 1. −2(a+c)=10 2. a^2 +4ac+b^2 +c^2 −d^2 =0 3. −2(a(c^2 −d^2 )+c(a^2 +b^2 ))=6 4. (a^2 +b^2 )(c^2 −d^2 )=−1 1. a=−c−5e 2. b=(√(−a^2 +4ac−c^2 +d^2 ))=(√(2c^2 +10c+d^2 −25)) 3. d=(√(ac+c^2 +((b^2 c+3)/a)))=(√(−((2c^3 +15c^2 +3)/(2c+5)))) 2. b=(√((2c^3 +15c^2 −128)/(2c+5))) 4. 4c^4 +40c^3 +75c^2 −125c+((689)/4)−((17161)/(4(2c+5)^2 ))=0 64(c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1)=0 c^6 +15c^5 +75c^4 +125c^3 +4c^2 +20c+1=0 c=u−((15)/6) u^6 −((75)/4)u^4 +((1939)/(16))u^2 −((17161)/(64))=0 u=(√v) v^3 −((75)/4)v^2 +((1939)/(16))v−((17161)/(64))=0 v=w+((75)/(12)) w^3 +4w+1=0 w=((−(q/2)+(√((p^3 /(27))+(q^2 /4)))))^(1/3) +((−(q/2)−(√((p^3 /(27))+(q^2 /4)))))^(1/3) = =((−(1/2)+((√(849))/(18))))^(1/3) +((−(1/2)−((√(849))/(18))))^(1/3) and now we have to go all the way back... not very friendly... as I mentioned before, it′s almost (or maybe absolutely) impossible to find a nicer exact form of w, so let me use the approximation w≈−.246266 v=((25)/4)+w≈6.00373 u=((√(25+4w))/2)≈2.45025 c=−(5/2)+((√(25+4w))/2)≈−.0497482 d=(√(((25)/2)−w−((131)/(2(√(25+4w))))))≈.787215i a=−(5/2)−((√(25+4w))/2)≈−4.95025 b=(√(−((25)/2)+w+((131)/(2(√(25+4w))))))≈5.11001i x_1 =a+bi≈−10.0603 x_2 =a−bi≈.159762 x_3 =c+d≈−.0497482+.787215i x_4 =c−d≈−.0497482−.787215i

$${x}^{\mathrm{4}} +\mathrm{10}{x}^{\mathrm{3}} +\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{those}\:\mathrm{who}\:\mathrm{want}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution}... \\ $$$$ \\ $$$$\left({x}−{a}−{b}\mathrm{i}\right)\left({x}−{a}+{b}\mathrm{i}\right)\left({x}−{c}−{d}\right)\left({x}−{c}+{d}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −\mathrm{2}\left({a}+{c}\right){x}^{\mathrm{3}} +\left({a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right){x}^{\mathrm{2}} − \\ $$$$\:\:\:\:\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right){x}+\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{1}.\:−\mathrm{2}\left({a}+{c}\right)=\mathrm{10} \\ $$$$\mathrm{2}.\:{a}^{\mathrm{2}} +\mathrm{4}{ac}+{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{d}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{3}.\:−\mathrm{2}\left({a}\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\right)=\mathrm{6} \\ $$$$\mathrm{4}.\:\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left({c}^{\mathrm{2}} −{d}^{\mathrm{2}} \right)=−\mathrm{1} \\ $$$$ \\ $$$$\mathrm{1}.\:{a}=−{c}−\mathrm{5}{e} \\ $$$$\mathrm{2}.\:{b}=\sqrt{−{a}^{\mathrm{2}} +\mathrm{4}{ac}−{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }=\sqrt{\mathrm{2}{c}^{\mathrm{2}} +\mathrm{10}{c}+{d}^{\mathrm{2}} −\mathrm{25}} \\ $$$$\mathrm{3}.\:{d}=\sqrt{{ac}+{c}^{\mathrm{2}} +\frac{{b}^{\mathrm{2}} {c}+\mathrm{3}}{{a}}}=\sqrt{−\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{2}.\:{b}=\sqrt{\frac{\mathrm{2}{c}^{\mathrm{3}} +\mathrm{15}{c}^{\mathrm{2}} −\mathrm{128}}{\mathrm{2}{c}+\mathrm{5}}} \\ $$$$\mathrm{4}. \\ $$$$\mathrm{4}{c}^{\mathrm{4}} +\mathrm{40}{c}^{\mathrm{3}} +\mathrm{75}{c}^{\mathrm{2}} −\mathrm{125}{c}+\frac{\mathrm{689}}{\mathrm{4}}−\frac{\mathrm{17161}}{\mathrm{4}\left(\mathrm{2}{c}+\mathrm{5}\right)^{\mathrm{2}} }=\mathrm{0} \\ $$$$\mathrm{64}\left({c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}\right)=\mathrm{0} \\ $$$${c}^{\mathrm{6}} +\mathrm{15}{c}^{\mathrm{5}} +\mathrm{75}{c}^{\mathrm{4}} +\mathrm{125}{c}^{\mathrm{3}} +\mathrm{4}{c}^{\mathrm{2}} +\mathrm{20}{c}+\mathrm{1}=\mathrm{0} \\ $$$${c}={u}−\frac{\mathrm{15}}{\mathrm{6}} \\ $$$${u}^{\mathrm{6}} −\frac{\mathrm{75}}{\mathrm{4}}{u}^{\mathrm{4}} +\frac{\mathrm{1939}}{\mathrm{16}}{u}^{\mathrm{2}} −\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${u}=\sqrt{{v}} \\ $$$${v}^{\mathrm{3}} −\frac{\mathrm{75}}{\mathrm{4}}{v}^{\mathrm{2}} +\frac{\mathrm{1939}}{\mathrm{16}}{v}−\frac{\mathrm{17161}}{\mathrm{64}}=\mathrm{0} \\ $$$${v}={w}+\frac{\mathrm{75}}{\mathrm{12}} \\ $$$${w}^{\mathrm{3}} +\mathrm{4}{w}+\mathrm{1}=\mathrm{0} \\ $$$${w}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}+\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}−\sqrt{\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}}}= \\ $$$$=\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{849}}}{\mathrm{18}}} \\ $$$$ \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{go}\:\mathrm{all}\:\mathrm{the}\:\mathrm{way}\:\mathrm{back}... \\ $$$$\mathrm{not}\:\mathrm{very}\:\mathrm{friendly}... \\ $$$$\mathrm{as}\:\mathrm{I}\:\mathrm{mentioned}\:\mathrm{before},\:\mathrm{it}'\mathrm{s}\:\mathrm{almost}\:\left(\mathrm{or}\:\mathrm{maybe}\right. \\ $$$$\left.\mathrm{absolutely}\right)\:\mathrm{impossible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{nicer}\:\mathrm{exact} \\ $$$$\mathrm{form}\:\mathrm{of}\:{w},\:\mathrm{so}\:\mathrm{let}\:\mathrm{me}\:\mathrm{use}\:\mathrm{the}\:\mathrm{approximation} \\ $$$$ \\ $$$${w}\approx−.\mathrm{246266} \\ $$$${v}=\frac{\mathrm{25}}{\mathrm{4}}+{w}\approx\mathrm{6}.\mathrm{00373} \\ $$$${u}=\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx\mathrm{2}.\mathrm{45025} \\ $$$${c}=−\frac{\mathrm{5}}{\mathrm{2}}+\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−.\mathrm{0497482} \\ $$$${d}=\sqrt{\frac{\mathrm{25}}{\mathrm{2}}−{w}−\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx.\mathrm{787215i} \\ $$$${a}=−\frac{\mathrm{5}}{\mathrm{2}}−\frac{\sqrt{\mathrm{25}+\mathrm{4}{w}}}{\mathrm{2}}\approx−\mathrm{4}.\mathrm{95025} \\ $$$${b}=\sqrt{−\frac{\mathrm{25}}{\mathrm{2}}+{w}+\frac{\mathrm{131}}{\mathrm{2}\sqrt{\mathrm{25}+\mathrm{4}{w}}}}\approx\mathrm{5}.\mathrm{11001i} \\ $$$${x}_{\mathrm{1}} ={a}+{b}\mathrm{i}\approx−\mathrm{10}.\mathrm{0603} \\ $$$${x}_{\mathrm{2}} ={a}−{b}\mathrm{i}\approx.\mathrm{159762} \\ $$$${x}_{\mathrm{3}} ={c}+{d}\approx−.\mathrm{0497482}+.\mathrm{787215i} \\ $$$${x}_{\mathrm{4}} ={c}−{d}\approx−.\mathrm{0497482}−.\mathrm{787215i} \\ $$

Question Number 36154 Answers: 0 Comments: 0

Q. If x≠y≠z and determinant ((x,x^3 ,(x^4 −1)),(y,y^3 ,(y^4 −1)),((z ),z^3 ,(z^4 −1)))=0 Prove that xyz(xy+yz+zx)=(x+y+z) please help.

$${Q}.\:\:{If}\:{x}\neq{y}\neq{z}\:\:{and}\:\:\begin{vmatrix}{{x}}&{{x}^{\mathrm{3}} }&{{x}^{\mathrm{4}} −\mathrm{1}}\\{{y}}&{{y}^{\mathrm{3}} }&{{y}^{\mathrm{4}} −\mathrm{1}}\\{{z}\:}&{{z}^{\mathrm{3}} }&{{z}^{\mathrm{4}} −\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$ \\ $$$${Prove}\:{that}\:\:{xyz}\left({xy}+{yz}+{zx}\right)=\left({x}+{y}+{z}\right) \\ $$$$ \\ $$$${please}\:{help}. \\ $$

Question Number 36153 Answers: 0 Comments: 1

(((x+yi−2)^2 )/(x−yi+1))

$$\frac{\left({x}+{yi}−\mathrm{2}\right)^{\mathrm{2}} }{{x}−{yi}+\mathrm{1}} \\ $$

Question Number 36132 Answers: 0 Comments: 7

a+b=10.........(i) ab+c=0..........(ii) ac+d=6..........(iii) ad=−1...........(iv) (a,b,c,d)=? Note: This problem is related to solve the equation (t^4 +10t+6t−1=0) of Q#35844

$${a}+{b}=\mathrm{10}.........\left(\mathrm{i}\right) \\ $$$${ab}+{c}=\mathrm{0}..........\left(\mathrm{ii}\right) \\ $$$${ac}+{d}=\mathrm{6}..........\left(\mathrm{iii}\right) \\ $$$${ad}=−\mathrm{1}...........\left(\mathrm{iv}\right) \\ $$$$\left({a},{b},{c},{d}\right)=? \\ $$$$\mathcal{N}{ote}:\:{This}\:{problem}\:{is}\:{related}\:{to}\:{solve} \\ $$$${the}\:{equation}\:\left({t}^{\mathrm{4}} +\mathrm{10}{t}+\mathrm{6}{t}−\mathrm{1}=\mathrm{0}\right)\:{of} \\ $$$${Q}#\mathrm{35844} \\ $$

Question Number 36126 Answers: 0 Comments: 4

x^4 +10x^3 +6x−1 =^(?) (x^2 +(((√5)−1)/2))(x^2 +10x−(((√5)+1)/2))

$$\mathrm{x}^{\mathrm{4}} +\mathrm{10x}^{\mathrm{3}} +\mathrm{6x}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\overset{?} {=}\left({x}^{\mathrm{2}} +\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} +\mathrm{10}{x}−\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}\right) \\ $$

Question Number 36115 Answers: 0 Comments: 1