SAT Nonlinear Equations & Systems Question 132 | Free SAT Prep | Aniko

00:00

Question 132·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

In the system of equations

\begin{cases} y = x^2 + kx + 4 \ y = -x^2 + 6x + 1 \end{cases}

$k$ is a real constant chosen so that the system has exactly one real solution.

What is the sum of all such values of $k$ ?

Your answer

(Express the answer as an integer)

For systems where both equations are polynomials in $x$ (often quadratics) and a parameter appears, equate the right-hand sides to get a single equation in $x$ . This equation’s real solutions match the intersection $x$ -values, so control the number of real solutions using the discriminant $b^2 - 4ac$ : set it to zero for exactly one solution, positive for two, or negative for none. After solving for the parameter values that meet the condition, carefully follow what the question asks—such as summing all valid parameter values—rather than stopping once you find them.

Hints

Click me!

Relate the system to a single equation in x

The solutions of the system are points where the two parabolas intersect. How can you combine the two equations into one equation that $x$ must satisfy at those intersection points?

Recognize the quadratic and think about number of solutions

After you set the two right-hand sides equal and simplify, you will get a quadratic equation in $x$ . What condition must hold for this quadratic to have exactly one real solution instead of two or none?

Use the discriminant

For a quadratic $ax^2 + bx + c = 0$ , the discriminant is $b^2 - 4ac$ . Set this equal to $0$ for exactly one real solution, then solve the resulting equation for $k$ , and finally add all the possible $k$ -values.

Desmos Guide

Set up the two parabolas with a slider for k

In Desmos, type y = x^2 + kx + 4 and y = -x^2 + 6x + 1. Desmos will automatically create a slider for k, which you can move to see how the number of intersection points changes as k changes.

Find values of k that give exactly one intersection point

Move the k slider and watch the graphs. You are looking for positions of k where the two parabolas touch at exactly one point (are tangent) instead of crossing at two points or not touching at all. Note the approximate values of k where this happens.

Confirm algebraic values and compute the sum

Once you have approximate k-values from the graph, you can verify them by typing the equation of the discriminant into Desmos, for example (k - 6)^2 = 24, and checking that the solutions match your slider values. Then, add these $k$ -values (either in Desmos or by hand) to get the requested sum.

Step-by-step Explanation

Set the equations equal to find intersection points

The solutions of the system are the intersection points of the two parabolas, so set the right-hand sides equal:

x^2 + kx + 4 = -x^2 + 6x + 1.

Move everything to one side:

\begin{aligned} x^2 + kx + 4 + x^2 - 6x - 1 &= 0 \\ 2x^2 + (k - 6)x + 3 &= 0. \end{aligned}

So the $x$ -coordinates of the intersection points must satisfy the quadratic

2x^2 + (k - 6)x + 3 = 0.

Use the discriminant condition for exactly one real solution

A quadratic $ax^2 + bx + c = 0$ has:

two distinct real solutions if its discriminant $b^2 - 4ac > 0$ ,
exactly one real solution (a double root) if $b^2 - 4ac = 0$ ,
no real solutions if $b^2 - 4ac < 0$ .

Here, $a = 2$ , $b = k - 6$ , and $c = 3$ . For the system to have exactly one real solution, we need

(k - 6)^2 - 4(2)(3) = 0.

Solve for all possible values of k

Now solve the equation from the discriminant:

\begin{aligned} (k - 6)^2 - 24 &= 0 \\ (k - 6)^2 &= 24 \\ k - 6 &= \pm\sqrt{24} \\ k - 6 &= \pm 2\sqrt{6}. \end{aligned}

So the possible values of $k$ are

\begin{aligned} k &= 6 + 2\sqrt{6}, \\ k &= 6 - 2\sqrt{6}. \end{aligned}

Find the required sum of all such k-values

The problem asks for the sum of all values of $k$ that make the system have exactly one real solution. Add the two values you found:

\begin{aligned} (6 + 2\sqrt{6}) + (6 - 2\sqrt{6}) &= 6 + 6 + 2\sqrt{6} - 2\sqrt{6} \\ &= 12. \end{aligned}

So, the sum of all such values of $k$ is $12$ .

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation