If x^1+x^(−1)=5, what is the value of x^4+x^(−4)?

Official Answer

Direct attempts to solve for x in this problem will run into quadratics that don't factor and horrible non-integers that need to be raised to fourth powers. Instead, let's focus on manipulating the equation to solve for x^4 + x^−4 directly.

Given the similar structure of the given information, it seems reasonable to begin by squaring the equation x^1+x^−1=5. Be careful, though, not to simply square each term; exponents do not distribute over addition. Instead, recognize the special quadratic. We're looking at two terms added and then squared, so this expression fits the form (a+b)^2=a^2+2ab+b^2. Thus our result will be


(x^1+x^−1^2=5^2

(x^1)^2+2(x^1)(x^−1)+(x^−1)^2=25


x^2+2+x^−2=25


x^2+x^−2=23


Now just repeat the process of squaring both sides once more:


(x^2+x^−2)^2=23^2


x^4+2(x^2)(x^−2)+x^−4=23^2


x^4+2+x^−4=23^2


x^4+x^−4=23^2−2


And it's not even really necessary to calculate 23^2 (which turns out to be 529). 23^2 must end in a 9, so 23^2−2 must end in a 7, and the answer has to be A.

\(x + (\frac{1}{x}) = 5\)

Squaring both sides of the equation

\(x^2 + (\frac{1}{x})^2 + 2 = 5^2\)

\(x^2 + (\frac{1}{x})^2 = 23\)

Squaring both sides of the equation

\(x^4 + (\frac{1}{x})^4 + 2 = 23^2\)

\(x^4 + (\frac{1}{x})^4 = 529 - 2\)

\(x^4 + (\frac{1}{x})^4 = 527\)

Option A

Hi mekoner

there is a very simple formula used here

\((a+b)^2=a^2+b^2+2ab\)

now instead of \(a\) & \(b\) use \(x\) & \(\frac{1}{x}\) here

\(x^1 + x^{−1} = 5\)
\((x^1 + x^{−1})^2 = 5^2…x^2+x^{-2}+2=25…x^2+x^{-2}=23\)
\((x^2+x^{-2})^2=23^2…x^4+x^{-4}+2=529…x^4+x^{-4}=527\)

Answer (A)

why do we need to square it? In the response below your original post, it says that "it's reasonable" to do. Can you explain please? Thanks

Hi rnz

we are given \(x^1+x^{-1}\) and need to arrive at \(x^4+x^{-4}\). So squaring the original equation will raise it to power of \(2\) i.e. \(x^2+x^{-2}\) and on further squaring this expression we will reach our destination

Can someone please explain how we arrive at (x^2+1/X^2+2)?

(x+1/x)^2 = (x^2+1/X^2+2)?

\(x^1 + x^{(−1)} = 5\)

Or, \(x + \frac{1}{x} = 5\)

Or, \(x^2 + 2 + \frac{1}{x^2} = 25\) (Squaring both sides )

Or, \(x^2 + \frac{1}{x^2} = 23\)

Or, \(x^4 + \frac{1}{x^4} = 527\) (Squaring both sides ) , Hence Answer must be (A)

\(x^1 + x^{(−1)} = x + \frac{1}{x}\)

\(x^1 + x^{(−1)} = 5\)

\(x + \frac{1}{x} = 5\)

Square both sides:

\((x + \frac{1}{x})(x + \frac{1}{x}) = 25\)

\(x^2 + \frac{x}{x}+ \frac{x}{x} + \frac{1}{x^2} = 25\)

\(x^2 + 1 + 1 + \frac{1}{x^2} = 25\)

\(x^2 + \frac{1}{x^2} = 23\)

(Bonus if you spot that we squared the right side and then subtracted 2...if we do that again, we get the right answer.)

Square both sides again:

\((x^2 + \frac{1}{x^2})(x^2 + \frac{1}{x^2}) = 23^2\)

\(x^4 + \frac{x^2}{x^2} + \frac{x^2}{x^2} + \frac{1}{x^4} = 529\)

\(x^4 + 1 + 1 + \frac{1}{x^4} = 529\)

\(x^4 + \frac{1}{x^4} = 527\)

Answer choice A.