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If-x-2-3x-1-0-then-find-the-value-of-x-2-x-1-x-1-x-2-2-

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Question Number 191552 by MATHEMATICSAM last updated on 25/Apr/23
If x^2 − 3x + 1 = 0 then find the value of (x^2 + x + (1/x) + (1/x^2 ))^2
$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$
Answered by mr W last updated on 25/Apr/23
x^2 −3x+1=0 ⇒x+(1/x)=3 (x^2 +x+(1/x)+(1/x^2 ))^2 =(x+(1/x)+x^2 +(1/x^2 ))^2 =[(x+(1/x))+(x+(1/x))^2 −2]^2 =(3+3^2 −2)^2 =100 ✓
$${x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3} \\ $$$$\left({x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\left({x}+\frac{\mathrm{1}}{{x}}+{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\left[\left({x}+\frac{\mathrm{1}}{{x}}\right)+\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} −\mathrm{2}\right]^{\mathrm{2}} \\ $$$$=\left(\mathrm{3}+\mathrm{3}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$$=\mathrm{100}\:\checkmark \\ $$
Answered by Rasheed.Sindhi last updated on 26/Apr/23
AnOther Way... x^2 − 3x + 1 = 0⇒1=3x−x^2 =x(3−x) (x^2 + x + (1/x) + (1/x^2 ))^2 =(x^2 + x + ((x(3−x))/x) + ((x(3−x))/x^2 ))^2 =(x^2 + x + 3−x + ((3−x)/x))^2 =(x^2 + 3 + ((3−x)/x))^2 =(((x^3 +3x+3−x)/x))^2 =(((x^3 +2x+3×1)/x))^2 =(((x^3 +2x+3x(3−x))/x))^2 =(((x^3 +2x+9x−3x^2 )/x))^2 =(x^2 −3x+1+10)^2 =(0+10)^2 =100 ✓
$$\mathrm{AnOther}\:\mathrm{W}{ay}… \\ $$$${x}^{\mathrm{2}} \:−\:\mathrm{3}{x}\:+\:\mathrm{1}\:=\:\mathrm{0}\Rightarrow\mathrm{1}=\mathrm{3}{x}−{x}^{\mathrm{2}} ={x}\left(\mathrm{3}−{x}\right) \\ $$$$\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:+\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\frac{{x}\left(\mathrm{3}−{x}\right)}{{x}}\:+\:\frac{{x}\left(\mathrm{3}−{x}\right)}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{3}−{x}\:+\:\frac{\mathrm{3}−{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} \:\:+\:\mathrm{3}\:+\:\frac{\mathrm{3}−{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{x}^{\mathrm{3}} +\mathrm{3}{x}+\mathrm{3}−{x}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{3}×\mathrm{1}}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{3}{x}\left(\mathrm{3}−{x}\right)}{{x}}\right)^{\mathrm{2}} \\ $$$$=\left(\frac{{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{9}{x}−\mathrm{3}{x}^{\mathrm{2}} }{{x}}\right)^{\mathrm{2}} \\ $$$$=\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}+\mathrm{10}\right)^{\mathrm{2}} \\ $$$$=\left(\mathrm{0}+\mathrm{10}\right)^{\mathrm{2}} =\mathrm{100}\:\checkmark \\ $$

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