SAT Nonlinear Equations & Systems Question 198 | Free SAT Prep | Aniko

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

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For real parameter $p$ , consider the system

\begin{aligned} y &= x^2 + px + 13 \\ y &= 4x + 4 \end{aligned}

The system has exactly one real solution for certain values of $p$ . What is the sum of all such values of $p$ ?

For systems involving a line and a parabola with a parameter, the fastest SAT approach is to eliminate $y$ by setting the right-hand sides equal and rewriting the result as a quadratic in $x$ . Recognize that the phrase 'exactly one real solution' corresponds to a quadratic with discriminant $D = 0$ , not $D > 0$ or $D < 0$ . Compute $D = b^2 - 4ac$ in terms of the parameter, set it equal to zero, and solve that simple equation for the parameter. Finally, pay close attention to what the question asks for (such as the sum of parameter values) so you combine your solutions correctly and do not confuse sums with products or ignore negative signs.

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Combine the two equations into one

Since both equations equal $y$ , set $x^2 + px + 13$ equal to $4x + 4$ and rearrange to form a single equation in $x$ .

Relate the number of solutions to the discriminant

Once you have a quadratic in $x$ , think about how the discriminant $b^2 - 4ac$ tells you the number of real solutions. What must the discriminant equal for there to be exactly one real solution?

Write and solve the discriminant equation

For your quadratic in $x$ , identify $a$ , $b$ , and $c$ , compute $D = b^2 - 4ac$ , set $D$ equal to 0, and solve the resulting equation for $p$ . Then use those $p$ -values to answer the question about their sum.

Desmos Guide

Graph the line and the parabola with a slider for p

In Desmos, type y1 = x^2 + p x + 13. When Desmos offers to create a slider for p, accept it. Then type y2 = 4x + 4 to graph the line.

Use the slider to find when there is exactly one intersection

Move the slider for p slowly and watch how many intersection points the line and parabola have. Find the values of p where the line just touches the parabola at a single point (the graphs are tangent) instead of crossing it twice or not at all.

Record and combine the p-values

Note the two distinct values of p for which there is exactly one intersection point. Then, outside of Desmos, add those two values together to get the sum requested in the problem.

Step-by-step Explanation

Set the equations equal to eliminate y

The solutions to the system are the intersection points of the line and the parabola, so set the right-hand sides equal:

x^2 + px + 13 = 4x + 4

Move all terms to one side to get a quadratic in $x$ :

\begin{aligned} x^2 + px + 13 - 4x - 4 &= 0 \\ x^2 + (p - 4)x + 9 &= 0. \end{aligned}

Now you have a quadratic equation in $x$ whose coefficients depend on $p$ . Each real solution $x$ of this quadratic corresponds to a real solution of the system.

Use the discriminant condition for exactly one real solution

For a quadratic $ax^2 + bx + c = 0$ , the discriminant is $D = b^2 - 4ac$ :

If $D > 0$ , there are 2 distinct real solutions.
If $D = 0$ , there is exactly 1 real solution.
If $D < 0$ , there are no real solutions.

Here, $a = 1$ , $b = p - 4$ , and $c = 9$ , so the discriminant is

D = (p - 4)^2 - 4 \cdot 1 \cdot 9 = (p - 4)^2 - 36.

We want exactly one real solution, so set $D = 0$ . This gives the equation

(p - 4)^2 - 36 = 0.

Solve for the parameter p

Solve the equation

(p - 4)^2 - 36 = 0.

Add 36 to both sides:

(p - 4)^2 = 36.

Take square roots of both sides:

p - 4 = \pm 6.

So there are two possible values of $p$ :

p = 4 + 6 \quad \text{and} \quad p = 4 - 6,

which means $p = 10$ and $p = -2$ are the values that make the system have exactly one real solution.

Find the required sum

The question asks for the sum of all such values of $p$ .

We found that $p = 10$ and $p = -2$ both work, so their sum is

10 + (-2) = 8.

Therefore, the correct answer is $8$ , which corresponds to choice B.

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Step-by-step Explanation

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Step-by-step Explanation