Question
If $x^2 + \frac{1}{x^2} = 16$ and $x \neq 0$, then what is the value of $x^4 + \frac{1}{x^4}$?
The correct answer is
254
Solving for $x^4 + \frac{1}{x^4}$
We are given the equation: $x^2 + \frac{1}{x^2} = 16$ Our goal is to find the value of $x^4 + \frac{1}{x^4}$.
Using Algebraic Identities
Notice that $x^4$ is the square of $x^2$, and $\frac{1}{x^4}$ is the square of $\frac{1}{x^2}$. We can use the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$.
Let $a = x^2$ and $b = \frac{1}{x^2}$. Squaring both sides of the given equation:
$ \left( x^2 + \frac{1}{x^2} \right)^2 = 16^2 $Expand the left side:
$ (x^2)^2 + 2(x^2)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x^2}\right)^2 = 256 $Simplify the terms:
$ x^4 + 2(1) + \frac{1}{x^4} = 256 $ $ x^4 + 2 + \frac{1}{x^4} = 256 $Final Calculation
To find the value of $x^4 + \frac{1}{x^4}$, subtract 2 from both sides of the equation:
$ x^4 + \frac{1}{x^4} = 256 - 2 $ $ x^4 + \frac{1}{x^4} = 254 $Therefore, the value of $x^4 + \frac{1}{x^4}$ is 254.
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Important Questions from Algebra (Notes)
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- In the expansion of (x + 9)(x - 6)(x + 5), what is the coefficient of x?
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- The roots of the equation $ax^3-24x^2+188x-480=0$ are three consecutive even natural numbers. The value of a is _____.
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