Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 217730 by ArshadS last updated on 19/Mar/25

$$x+\frac{1}{x}=3;x^5+\frac{1}{x^5}=?$$

Answered by Wuji last updated on 19/Mar/25

$$let{S}_n=x^n+\frac{1}{x^n}$$$$S_1=x+\frac{1}{x}=3$$$$S_2={(x+\frac{1}{x})}^2-2\Rightarrow S_2=3^2-2=7$$$$S_3=(x+\frac{1}{x})S_2-S_1\Rightarrow S_3=3\times 7-3=18$$$$S_4=(x+\frac{1}{x})S_3-S_2\Rightarrow S_4=3\times 18-7=47$$$$S_5=(x+\frac{1}{x})S_4-S_3\Rightarrow S_5=3\times 47-18=123$$$$S_5=x^5+\frac{1}{x^5}=123$$

Commented by ArshadS last updated on 22/Mar/25

$$Thankssir!$$

Answered by Rasheed.Sindhi last updated on 19/Mar/25

$$x+\frac{1}{x}=3;x^5+\frac{1}{x^5}=?$$$$x+\frac{1}{x}=3\Rightarrow x^2=3x-1$$$$\Rightarrow x^4=9x^2-6x+1$$$$=9(3x-1)-6x+1$$$$=21x-8$$$$\Rightarrow \overline{)\begin{array}{c}{x}^5\end{array}}=21x^2-8x=21(3x-1)-8x$$$$=55x-21$$$$$$$$\frac{1}{x}=3-x$$$$\frac{1}{x^2}={(3-x)}^2=x^2-6x+9$$$$=(3x-1)-6x+9=-3x+8$$$$\frac{1}{x^4}={(-3x+8)}^2=9x^2-48x+64$$$$=9(3x-1)-48x+64$$$$=27x-9-48x+64$$$$=-21x+55$$$$\overline{)\begin{array}{c}\frac{1}{x^5}\end{array}}=-21+55\left(\frac{1}{x}\right)=-21+55(3-x)$$$$=-21+165-55x=-55x+144$$$$$$$$\overline{)\begin{array}{c}{x}^5+\frac{1}{x^5}\end{array}}=(55x-21)+(-55x+144)$$$$=123$$

Commented by ArshadS last updated on 22/Mar/25

$$Thankssir!$$

Answered by Ghisom last updated on 20/Mar/25

$$x^2-3x+1=0$$$$$$$$x^2=3x-1$$$$x^5=x{\left(x^2\right)}^2=x{(3x-1)}^2=x(9x^2-6x+1)=$$$$=x(9(3x-1)-6x+1)=x(21x-8)=$$$$=21x^2-8x=21(3x-1)-8x=55x-21$$$$$$$$\frac{1}{x^5}=\frac{1}{55x-21}=\frac{1}{55(3-\frac{1}{x})-21}=\frac{x}{144x-55}$$$$$$$$55x-21+\frac{x}{144x-55}=$$$$=\frac{(55x-21)(144x-55)+x}{144x-55}=$$$$=\frac{7920x^2-6048x+1155}{144x-55}=$$$$=\frac{7920(3x-1)-6048x+1155}{144x-55}=$$$$=\frac{17712x-6765}{144x-55}=123$$

Commented by ArshadS last updated on 22/Mar/25

$$Thankssir!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com