Algebraic Value Calculation
The question asks to find the value of $x^{2} + \frac{1}{x^{2}}$ given the initial condition $x + \frac{1}{x} = 2$. This can be solved by squaring the given equation.
Solving for $x^{2} + \frac{1}{x^{2}}$
Step 1: Write down the given equation.
The provided equation is:
$x + \frac{1}{x} = 2$
Step 2: Square both sides.
Squaring both sides of the equation allows us to introduce the terms $x^2$ and $\frac{1}{x^2}$.
$ \left(x + \frac{1}{x}\right)^{2} = (2)^{2} $
Step 3: Expand the left side.
Use 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} = 4 $
Step 4: Simplify the expanded equation.
The middle term $2(x)\left(\frac{1}{x}\right)$ simplifies to $2$.
$ x^{2} + 2 + \frac{1}{x^{2}} = 4 $
Step 5: Isolate the desired term.
Subtract 2 from both sides to find the value of $x^{2} + \frac{1}{x^{2}}$.
$ x^{2} + \frac{1}{x^{2}} = 4 - 2 $
Step 6: Final result.
$ x^{2} + \frac{1}{x^{2}} = 2 $
The value of the expression $x^{2} + \frac{1}{x^{2}}$ is 2.
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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:
