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If \(x + \frac{1}{x} = 3\) then, what is the value of \(x^5 + \frac{1}{x^5}\) ?
A. 123
B. 144
C. 159
D. 186
E. 191
Are You Up For the Challenge: 700 Level Questions
A. 123
B. 144
C. 159
D. 186
E. 191
Are You Up For the Challenge: 700 Level Questions
02 Mar 2020, 06:52
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\(x + \frac{1}{x} = 3\)
Squaring both sides
x^2 + 1/x^2 = 7, Again
Squaring both sides
x^4 + 1/x^4 = 49-2 = 47. Multiplying both sides by \(x + \frac{1}{x} \)
(x^4 + 1/x^4 ) + (x + 1/x) = 47*3 = 141
x^5 + x^3 + 1/x^3 + (1/x^5) = 141
Let's take \(x + \frac{1}{x} = 3\) again, cubing both sides
x^3 + (1/x)^3 + 3(3) = 27
x^3 + (1/x)^3 = 27-9=18.
so, x^5 + (1/x^5) = 141 -18= 123
Squaring both sides
x^2 + 1/x^2 = 7, Again
Squaring both sides
x^4 + 1/x^4 = 49-2 = 47. Multiplying both sides by \(x + \frac{1}{x} \)
(x^4 + 1/x^4 ) + (x + 1/x) = 47*3 = 141
x^5 + x^3 + 1/x^3 + (1/x^5) = 141
Let's take \(x + \frac{1}{x} = 3\) again, cubing both sides
x^3 + (1/x)^3 + 3(3) = 27
x^3 + (1/x)^3 = 27-9=18.
so, x^5 + (1/x^5) = 141 -18= 123
Bunuel
13 Jul 2021, 20:16
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Bunuel
Step I: Square the equation
\((x + \frac{1}{x})^2 = 3^2\)
\(x ^2+ \frac{1}{x^2} +2= 9\)
\(x ^2+ \frac{1}{x^2} = 7\)
Step I: Cube the equation
\((x + \frac{1}{x})^3 = 3^3\)
\(x ^3+ \frac{1}{x^3} +3x*\frac{1}{x})(x+\frac{1}{x})= 27\)
\(x ^3+ \frac{1}{x^3} +3*3= 27\)
So, \(x ^3+ \frac{1}{x^3} = 18\)
Multiply the two equations that you get from the two steps.
\((x ^3+ \frac{1}{x^3})(x^2+\frac{1}{x^2})= 7*18=126\)
\(x ^5+ \frac{1}{x^5}+\frac{x^2}{x^3}+\frac{x^3}{x^2})=126\)
\((x ^5+ \frac{1}{x^5})+(x+\frac{1}{x})=126\)
\((x ^5+ \frac{1}{x^5})+(3)=126\)
\((x ^5+ \frac{1}{x^5})=123\)
A
