Question 69·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Type $(x_1-2)^4-20(x_1-2)^2+64\sim0\{x_1>5\}$ in Desmos. In a regression, $x_1$ is the value Desmos finds, $\sim$ tells it to match the two sides, and the braces enforce greater than $5$. Under PARAMETERS, Desmos reports $x_1=6$. So the solution the prompt wants is $x=6$.

Put the allowed solution $x=6$ into the given form $x=2+\sqrt{k}$: $6=2+\sqrt{k}$. Subtract $2$: $4=\sqrt{k}$. The $4$ is the square root of $k$, not yet $k$.

Square both sides to find $k$: $4^2=k$. Stopping at $4$ would answer with $\sqrt{k}$ instead.

Type $4^2$ on a new Desmos line. It displays $16$. Finding $x$ is a stepping stone; finish with the quantity the question asks for. The positive integer in $x=2+\sqrt{k}$ is $k=16$. Choice D.

Use a substitution, a short name for the repeated square: $u=(x-2)^2$. Then $(x-2)^4=u^2$, so the equation becomes:

Find two numbers that multiply to $64$ and add to $-20$: $-4$ and $-16$. Factor:

The zero-product property says a product is zero only if at least one factor is zero. Set each factor to zero: $u-4=0\quad\text{or}\quad u-16=0$. Solve those equations:

The required $x$ is greater than $5$. Subtract $2$: $x-2>3$. Square both positive sides: $u=(x-2)^2>9$. So $u=4$ is excluded, leaving $u=16$.

From the given form $x=2+\sqrt{k}$, subtract $2$: $x-2=\sqrt{k}$. Square both sides: $u=(x-2)^2=k$. Since the allowed value is $u=16$, the positive integer is $k=16$. Choice D.