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

Answer: (D)

$7+3\sqrt{5}$

Answer 2

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 $7+3\sqrt{5}​.$