In the system of equations above, is a constant. The system has exactly one solution . What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 168 | Free SAT Prep | Aniko

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Question 168·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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\begin{aligned} y &= x^2 + 5x + k \\ y &= x + 7 \end{aligned}

In the system of equations above, $k$ is a constant. The system has exactly one solution $(x,y)$ . What is the value of $k$ ?

When a system involves a line and a parabola and asks for conditions that give a specific number of solutions, first set the equations equal to turn it into a single quadratic in one variable. Then use the discriminant $b^2 - 4ac$ to control the number of real solutions: positive for two, zero for exactly one, and negative for none. Setting the discriminant to zero and solving for the parameter (here, $k$ ) is usually the fastest and most reliable approach on the SAT.

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Use substitution between the two equations

Both equations are equal to $y$ . What equation do you get if you set $x^2 + 5x + k$ equal to $x + 7$ and simplify?

Connect intersections to quadratic solutions

After you rewrite the system as a single quadratic equation in $x$ , think about how the number of solutions to that quadratic relates to how many intersection points the line and the parabola have.

Apply the discriminant condition

For a quadratic $ax^2+bx+c=0$ , recall the discriminant $b^2-4ac$ . What must this expression equal if the quadratic is to have exactly one real solution? Use this to form an equation involving $k$ and solve it.

Desmos Guide

Graph the parabola with a slider for k

In Desmos, enter y = x^2 + 5x + k. When prompted, add a slider for k so you can change its value and see how the parabola moves up and down.

Graph the line

Enter the second equation y = x + 7 in Desmos. You will now see both the line and the parabola on the same coordinate plane.

Adjust k until there is exactly one intersection

Move the k slider and watch how the number of intersection points between the line and the parabola changes. Find the value of k for which the graphs just touch at a single point (are tangent). Read this k value from the slider—that is the value that makes the system have exactly one solution.

Step-by-step Explanation

Set the equations equal

Because both equations equal $y$ , set their right-hand sides equal:

x^2 + 5x + k = x + 7

Move everything to one side to make a quadratic equation in $x$ :

x^2 + 5x + k - x - 7 = 0 \\[0.7em] x^2 + 4x + (k - 7) = 0

So the $x$ -values where the graphs intersect must satisfy $x^2 + 4x + (k - 7) = 0$ . This is where the number of solutions is determined.

Interpret “exactly one solution”

The system has exactly one solution $(x,y)$ when the line and the parabola intersect at exactly one point.

For the quadratic equation $x^2 + 4x + (k - 7) = 0$ :

Two distinct real solutions mean two intersection points.
Zero real solutions mean no intersection.
Exactly one real solution means the parabola and line just touch (are tangent), so the quadratic has a double root, which happens when its discriminant is $0$ .

Write the discriminant and set it to zero

For the quadratic $x^2 + 4x + (k - 7) = 0$ :

$a = 1$
$b = 4$
$c = k - 7$

The discriminant is $b^2 - 4ac$ .

Compute it in terms of $k$ :

\Delta = 4^2 - 4(1)(k - 7) = 16 - 4(k - 7)

For exactly one real solution, set $\Delta = 0$ :

16 - 4(k - 7) = 0

Now solve this equation for $k$ .

Solve for the value of k

Solve

16 - 4(k - 7) = 0

First distribute the $-4$ :

16 - 4k + 28 = 0 \\[0.7em] 44 - 4k = 0

Add $4k$ to both sides:

44 = 4k

Divide both sides by $4$ :

k = 11

So the value of $k$ that makes the system have exactly one solution is $11$ , which corresponds to choice A.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation