$16(x^4 + \frac{1}{x^4}) - 257 = 0$
\(\frac{17}{4}\)
To solve the problem, we need to find the value of \(x^2 + \frac{1}{x^2}\) given the equation:
\(16(x^4 + \frac{1}{x^4}) - 257 = 0.\)
Let's proceed step-by-step:
- \(16(x^4 + \frac{1}{x^4}) = 257\)
- Simplifying, we get:
- \(x^4 + \frac{1}{x^4} = \frac{257}{16}\)
- Recall the identity:
- \((x^2 + \frac{1}{x^2})^2 = x^4 + \frac{1}{x^4} + 2\)
- Now, substitute the value we found:
- \((x^2 + \frac{1}{x^2})^2 = \frac{257}{16} + 2\)
- Convert and simplify:
- \(\frac{257}{16} + 2 = \frac{257}{16} + \frac{32}{16} = \frac{257 + 32}{16} = \frac{289}{16}\)
- Thus,
- \((x^2 + \frac{1}{x^2})^2 = \frac{289}{16}\)
- Taking the square root on both sides gives:
- \(x^2 + \frac{1}{x^2} = \sqrt{\frac{289}{16}} = \frac{\sqrt{289}}{\sqrt{16}} = \frac{17}{4}\)
Finally, the correct answer to the problem according to computations and given options is \(\frac{17}{4}\)
Important Questions from Algebra
The difference between the two positive numbers x and y where x > y, is 25% of x. If the value of y is 15, then the value of x is:
If p2 + q2 - r2 = 0, then the value of (p6 + q6 - r6) ÷ p2q2r2 is:
If √2 + √x = √3, then the value of x is equal to:
The sum of two numbers is 20 and their difference is 2.5. Ratio of these numbers will be:
If \(\rm \frac{\sqrt{19 - x \sqrt{12}}}{1} = \sqrt 4 - \sqrt 3\) then the value of x is equal to:
