Algebra: Finding $x^2 + \frac{1}{x^2}$ from $x + \frac{1}{x}$
We are given the equation $x + \frac{1}{x} = 4$. Our goal is to find the value of $x^2 + \frac{1}{x^2}$.
Solving for $x^2 + \frac{1}{x^2}$
To find the value of $x^2 + \frac{1}{x^2}$, we can square both sides of the given equation:
$ \left(x + \frac{1}{x}\right)^2 = 4^2 $
Expand the left side using the algebraic identity $(a+b)^2 = a^2 + 2ab + b^2$. Here, $a=x$ and $b=\frac{1}{x}$:
$ x^2 + 2(x)\left(\frac{1}{x}\right) + \left(\frac{1}{x}\right)^2 = 16 $
Simplify the middle term. Note that $x \times \frac{1}{x} = 1$:
$ x^2 + 2(1) + \frac{1}{x^2} = 16 $
$ x^2 + 2 + \frac{1}{x^2} = 16 $
Now, isolate the term $x^2 + \frac{1}{x^2}$ by subtracting 2 from both sides:
$ x^2 + \frac{1}{x^2} = 16 - 2 $
$ x^2 + \frac{1}{x^2} = 14 $
Therefore, the value of $x^2 + \frac{1}{x^2}$ is 14.
Similar Questions
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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).
