SAT Question 13e57f0a: Nonlinear Equations In One Variable And Systems Of Equations In Two Variables answer and explanation with Desmos solution

Choice B is correct. Multiplying each side of the given equation by $-16$ yields $64x^2+112x=576$. To complete the square, adding $49$ to each side of this equation yields $64x^2+112x+49=576+49$, or ${\left(8x+7\right)}^2=625$. Taking the square root of each side of this equation yields two equations: $8x+7=25$ and $8x+7=-25$. Subtracting $7$ from each side of the equation $8x+7=25$ yields $8x=18$. Dividing each side of this equation by $8$ yields $x=\frac{18}{8}$, or $x=\frac{9}{4}$. Therefore, $\frac{9}{4}$ is a solution to the given equation. Subtracting $7$ from each side of the equation $8x+7=-25$ yields $8x=-32$. Dividing each side of this equation by $8$ yields $x=-4$. Therefore, the given equation has two solutions, $\frac{9}{4}$ and $-4$. Since $\frac{9}{4}$ is positive, it follows that $\frac{9}{4}$ is the positive solution to the given equation.

Alternate approach: Adding $4x^2$ and $7x$ to each side of the given equation yields $0=4x^2+7x-36$. The right-hand side of this equation can be rewritten as $4x^2+16x-9x-36$. Factoring out the common factor of $4x$ from the first two terms of this expression and the common factor of $-9$ from the second two terms yields $4x\left(x+4\right)-9\left(x+4\right)$. Factoring out the common factor of $\left(x+4\right)$ from these two terms yields the expression $\left(4x-9\right)\left(x+4\right)$. Since this expression is equal to $0$, it follows that either $4x-9=0$ or $x+4=0$. Adding $9$ to each side of the equation $4x-9=0$ yields $4x=9$. Dividing each side of this equation by $4$ yields $x=\frac{9}{4}$. Therefore, $\frac{9}{4}$ is a positive solution to the given equation. Subtracting $4$ from each side of the equation $x+4=0$ yields $x=-4$. Therefore, the given equation has two solutions, $\frac{9}{4}$ and $-4$. Since $\frac{9}{4}$ is positive, it follows that $\frac{9}{4}$ is the positive solution to the given equation.