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 \(x+1=x^2\) with \(x>0\), the objective is to determine the value of \(2x^4.\)
To find \(2x^4\), we first solve for \(x:\)
\(​x+1=x^2\)
\(⇒x^2−x−1=0\)
\(⇒x^2−x+\frac{1}{4}​=\frac{5}{4}​\)

\(⇒ \left(x−\frac{1}{2}​\right)^2=\frac{5}{4}​\)

\(⇒x−\frac{1}{2}​=±\sqrt{\frac{5}{4}}​​\)
\(⇒x=\frac{1}{2}​±\sqrt{\frac{5}{4}}\)​​​

Now, expressing \(2x^4:\)
\(​(x+1)^2=(x^2)^2\)
\(⇒x^2+2x+1=x^4\)
\(⇒3x+2=x^4\)
\(⇒6x+4=2x^4​\)

Substituting the found values of \(x=\frac{1}{2}​±\sqrt{\frac{5}{4}}​​:\)

\(6(1+\sqrt{\frac{5}{2}}​​)+4=2x^4\)
\(⇒3+3\sqrt{5}​+4=2x^4\)
\(⇒7+3\sqrt{5}​=2x^4​\)

Hence, if \(x+1=x^2\) and \(x>0\), then \(2x^4\) is equal to \(7+3\sqrt{5}​.\)