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
In the given question, two equations numbered l and II are given. Solve both the equations and mark the appropriate answer.
I. x2 – 26x + 165 = 0
II. y2 – 38y + 357 = 0
Factorize the following:
(x 2- 6xy + 9y 2) - 25
If P and Q are the points on the line Joining A(-2, 5) and B(3, 1) such that
AP = PQ = QB, then the mid point of PQ is
If a number and its reciprocal added it becomes 6, then what will be sum of its square and square of its reciprocal?
If 3x + 2y = 15, and xy = 6. Find the value of (3x3/2) + (4y3/9).
