Question-88141

Question Number 88141 by A8;15: last updated on 08/Apr/20

Answered by MJS last updated on 08/Apr/20

squaring and transforming 3 times leads to x^8 −28x^6 +238x^4 −588x^2 +128x−7=0 this can be factorized (x^2 +2x−7)(x^3 −x^2 −9x+1)^2 =0 x_(1, 2) =−1±2(√2) x_3 =(1/3)(1−4(√7)sin ((arcsin ((√7)/(14)))/3)) x_4 =(1/3)(1+4(√7)sin ((π+arcsin ((√7)/(14)))/3)) x_5 =(1/3)(1−4(√7)sin ((2π+arcsin ((√7)/(14)))/3)) testing all solutions leads to x=x_4

$$\mathrm{squaring}\:\mathrm{and}\:\mathrm{transforming}\:\mathrm{3}\:\mathrm{times}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}^{\mathrm{8}} −\mathrm{28}{x}^{\mathrm{6}} +\mathrm{238}{x}^{\mathrm{4}} −\mathrm{588}{x}^{\mathrm{2}} +\mathrm{128}{x}−\mathrm{7}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{can}\:\mathrm{be}\:\mathrm{factorized} \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{7}\right)\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}_{\mathrm{1},\:\mathrm{2}} =−\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${x}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{4}\sqrt{\mathrm{7}}\mathrm{sin}\:\frac{\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{4}\sqrt{\mathrm{7}}\mathrm{sin}\:\frac{\pi+\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}}{\mathrm{3}}\right) \\ $$$${x}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{4}\sqrt{\mathrm{7}}\mathrm{sin}\:\frac{\mathrm{2}\pi+\mathrm{arcsin}\:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}}{\mathrm{3}}\right) \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{solutions}\:\mathrm{leads}\:\mathrm{to}\:{x}={x}_{\mathrm{4}} \\ $$

Commented by ajfour last updated on 08/Apr/20

$${that}\:{it}\:{could}\:{be}\:{factorised},\:{how} \\ $$$${can}\:{one}\:{guess},\:{Sir}\:? \\ $$

Commented by MJS last updated on 08/Apr/20

I tried to approximate and found x_1 ≈−3.82843 and x_2 ≈1.82843 knowing that (√2)≈1.41421 it was easy to guess x_1 =^? −1−2(√2) and x_2 =^? −1+2(√2) it was just good luck

$$\mathrm{I}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{approximate}\:\mathrm{and}\:\mathrm{found} \\ $$$${x}_{\mathrm{1}} \approx−\mathrm{3}.\mathrm{82843}\:\mathrm{and}\:{x}_{\mathrm{2}} \approx\mathrm{1}.\mathrm{82843} \\ $$$$\mathrm{knowing}\:\mathrm{that}\:\sqrt{\mathrm{2}}\approx\mathrm{1}.\mathrm{41421}\:\mathrm{it}\:\mathrm{was}\:\mathrm{easy}\:\mathrm{to} \\ $$$$\mathrm{guess}\:{x}_{\mathrm{1}} \overset{?} {=}−\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}\:\mathrm{and}\:{x}_{\mathrm{2}} \overset{?} {=}−\mathrm{1}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{it}\:\mathrm{was}\:\mathrm{just}\:\mathrm{good}\:\mathrm{luck} \\ $$

Commented by ajfour last updated on 08/Apr/20

$${i}\:{see},\:{thank}\:{u}\:{Sir}. \\ $$

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