Question

If \(x + 1 = x^2\) and \(x > 0\), then \(2x^4\) is

• A. \(6+4\sqrt{5}\)
• B. \(3+5\sqrt{5}\)
• C. \(5+3\sqrt{5}\)
• D. \(7+3\sqrt{5}\)

Answer 1

The Correct Option is D

Given the equation \(x + 1 = x^2\) and \(x > 0\), we are to find the value of \(2x^4\). 

First, solve the equation:

\(x^2 - x - 1 = 0\)

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), with \(a = 1\), \(b = -1\), and \(c = -1\):

\(x = \frac{1 \pm \sqrt{1 + 4}}{2} = \frac{1 \pm \sqrt{5}}{2}\)

Since \(x > 0\), we take \(x = \frac{1 + \sqrt{5}}{2}\).

Next, calculate \(2x^4\):

First, find \(x^2\):

\(x^2 = (x+1) = \left(\frac{1 + \sqrt{5}}{2} + 1\right) = \frac{3 + \sqrt{5}}{2}\)

Now, find \(x^4\) by squaring \(x^2\):

\((x^2)^2 = \left(\frac{3 + \sqrt{5}}{2}\right)^2\)

Using \((a + b)^2 = a^2 + 2ab + b^2\), we have:

\(a = 3/2\), \(b = \sqrt{5}/2\)

\((x^2)^2 = \left(\frac{3}{2}\right)^2 + 2 \cdot \frac{3}{2} \cdot \frac{\sqrt{5}}{2} + \left(\frac{\sqrt{5}}{2}\right)^2\)

\((x^2)^2 = \frac{9}{4} + \frac{3\sqrt{5}}{2} + \frac{5}{4}\)

\((x^2)^2 = \frac{14}{4} + \frac{3\sqrt{5}}{2} = \frac{7}{2} + \frac{3\sqrt{5}}{2}\)

\(x^4 = \frac{7 + 3\sqrt{5}}{2}\)

Finally, compute \(2x^4\):

\(2x^4 = 2 \cdot \frac{7 + 3\sqrt{5}}{2}\)

\(2x^4 = 7 + 3\sqrt{5}\)

Thus, the value of \(2x^4\) is \(7 + 3\sqrt{5}\).