Question 223·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Since $x^{2/3}=(x^{1/3})^2$, the equation is a quadratic, an equation with a squared term, in $x^{1/3}$. A quadratic has at most two real values for that cube root. Cubing each value gives exactly one $x$, so there can be at most two real solutions.

Pick $15$ as a cutoff between the two smallest positive answer choices. Type the original equation with every $x$ changed to $x_1$, and change $=$ to $\sim$. The regression uses $\sim$ to find a value that makes the sides equal. Restrict it to $0<x_1<15$; Desmos shows $x_1=8$ under PARAMETERS, so $8$ is one solution.

Keep that line and type a second regression with $x_2$ so it fits a different value. Restrict this one to $x_2>15$; Desmos shows $x_2=27$ under PARAMETERS. You have found two roots, and the quadratic structure says there can be no more. Since $27>8$, the greater real solution is $27$. Choice C.

Let $u=x^{1/3}$, the cube root of $x$. Then $x^{2/3}=u^2$, so replace the two powers in the given equation:

Find two numbers that multiply to $6$ and add to $-5$: $-2$ and $-3$. Factor:

The zero-product property says a product is zero only if at least one factor is zero. Set each factor to zero:

Add $2$ or $3$, respectively:

The question asks for $x$, not $u$. Since $u=x^{1/3}$, cube both values to undo the cube root:

The greater real solution is $27$. Choice C.