Solving for $x^4 + \frac{1}{x^4}$
We are given the equation $x - \frac{1}{x} = 5$. We need to find the value of $x^4 + \frac{1}{x^4}$.
Step 1: Find $x^2 + \frac{1}{x^2}$
Square both sides of the given equation:
$ \left(x - \frac{1}{x}\right)^2 = 5^2 $
Expand the left side using the formula $(a-b)^2 = a^2 - 2ab + b^2$:
$ x^2 - 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 25 $
Simplify the equation:
$ x^2 - 2 + \frac{1}{x^2} = 25 $
Isolate $x^2 + \frac{1}{x^2}$:
$ x^2 + \frac{1}{x^2} = 25 + 2 $
$ x^2 + \frac{1}{x^2} = 27 $
Step 2: Find $x^4 + \frac{1}{x^4}$
Square both sides of the equation obtained in Step 1:
$ \left(x^2 + \frac{1}{x^2}\right)^2 = 27^2 $
Expand the left side using the formula $(a+b)^2 = a^2 + 2ab + b^2$:
$ (x^2)^2 + 2(x^2)\left(\frac{1}{x^2}\right) + \left(\frac{1}{x^2}\right)^2 = 729 $
Simplify the equation:
$ x^4 + 2 + \frac{1}{x^4} = 729 $
Isolate $x^4 + \frac{1}{x^4}$:
$ x^4 + \frac{1}{x^4} = 729 - 2 $
$ x^4 + \frac{1}{x^4} = 727 $
Final Answer Calculation
The value of $x^4 + \frac{1}{x^4}$ is 727.
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Important Questions from Algebra
If 2x – y = 2 and xy = \(\frac{3}{2}\) , then what is the value of x 3– \(\frac{{{y^3}}}{8}\) ?
If (10a 3+ 4b 3) : (11a 3- 15b 3) = 7 : 5, then (3a + 5b) : (9a - 2b) =?
- If 4sin 2 θ = 3(1+ cos θ), 0° < θ < 90°, then what is the value of (2tan θ + 4sin θ - sec θ)?
The value of:
\(\frac{{\sin 23^\circ \cos 67^\circ + \sec52^\circ \sin38^\circ + \cos 23^\circ \sin 67^\circ + \rm cosec52^\circ \cos 38^\circ }}{{\rm cose{c^2}20^\circ - {{\tan }^2}70^\circ }}\)
If (x + y) 3+ 27(x - y) 3= (Ax - 2y)(Bx 2+ Cxy + 13y 2), then the value of A - B - C is:
