Question 183·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Type $y=x+3$ and $x^2+y^2=25$ on separate Desmos lines. The graphs cross twice; each intersection satisfies both equations. Click the left crossing, about $(-4.702,-1.702)$. Its first coordinate is negative, so it isn't the point the question wants.

Click the right crossing, about $(1.702,4.702)$. Its $x$-coordinate is near $1.702$; $4.702$ is its $y$-coordinate. Type $\frac{3+\sqrt{41}}{2}$ and $\frac{-3+\sqrt{41}}{2}$. Desmos shows about $4.702$ and $1.702$, respectively. The other two expressions are negative because $\sqrt{41}>3$. So the positive value of $x$ is $\frac{-3+\sqrt{41}}{2}$. Choice D.

No Desmos needed here. Substitution keeps the square root exact. The line says $y=x+3$, so replace all of $y$ in the other equation with $x+3$:

Expand the square, including its middle term:

Combine the $x^2$ terms:

Subtract $25$ from both sides:

Divide every term by $2$:

For $ax^2+bx+c=0$, the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$. Here $a=1$, $b=3$, and $c=-8$. Substitute, keeping the negative sign on $c$:

Work out the expression under the square root:

Add inside the square root:

Since $41>9$, its square root is greater than $3$. So adding $\sqrt{41}$ to $-3$ gives a positive numerator; subtracting it gives a negative one. The positive value of $x$ is $\frac{-3+\sqrt{41}}{2}$. Choice D.