Question 198·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

No Desmos needed yet. First turn the system into an equation for an intersection. At a point on both graphs, the same $x$ gives the same $y$. Set the two expressions equal:

Subtract $4x+4$ to put the quadratic in zero form:

For $ax^2+bx+c=0$, the discriminant is $b^2-4ac$. Here $a=1$, $b=p-4$, and $c=9$. Exactly one real solution means the discriminant is zero: the two possible quadratic roots become the same root. Set it to zero:

Simplify:

Type $(p_1-4)^2\sim36\{p_1<4\}$. The $\sim$ asks Desmos to fit $p_1$; the braces keep the solution below $4$. Under PARAMETERS, Desmos shows $p_1=-2$. The subscript in $p_1$ lets Desmos solve for the unknown rather than make a slider for $p$.

Add $(p_2-4)^2\sim36\{p_2>4\}$. This time the braces keep the solution above $4$. Desmos shows $p_2=10$ under PARAMETERS. Squaring hides whether $p-4$ was positive or negative, so both values matter.

The question asks for the sum, not one value of $p$. Type $p_1+p_2$; Desmos prints $8$. So the sum of all values of $p$ that give exactly one real solution is $8$. Choice B.

Starting from $(p-4)^2=36$, expand the square:

Subtract $36$: $p^2-8p-20=0$. This form lets you find the sum without finding each value of $p$.

For $ap^2+bp+c=0$, the sum of the roots, meaning its two solutions for $p$, is $-b/a$. Here the $p$-coefficient is $-8$ and the $p^2$-coefficient is $1$, so their sum is $-(-8)/1=8$. Choice B.