The system of equations below involves the constant . The system has exactly two distinct real solutions. What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 30 | Free SAT Prep | Aniko

00:00

Question 30·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

The system of equations below involves the constant $m$ .

\begin{aligned} y &= x^3 - 4x \\ y &= mx + 2 \end{aligned}

The system has exactly two distinct real solutions. What is the value of $m$ ?

Your answer

(Express the answer as an integer)

For systems involving a polynomial and a line with a parameter, first eliminate $y$ to get a single equation in $x$ whose roots are the intersection $x$ -values. When the question asks for a specific number of real solutions, think in terms of root multiplicity: “exactly two distinct real solutions” for a cubic means one double root and one simple root. Enforce a double root by setting the polynomial and its derivative equal at the same $x$ -value, solve this small system to find the root and the parameter, and then (time permitting) quickly factor or check the resulting polynomial to confirm the number of distinct real roots matches the condition.

Hints

Click me!

Eliminate y first

Start by setting the two expressions for $y$ equal to each other so that you get one equation involving only $x$ , $m$ , and constants.

Think about what “exactly two distinct real solutions” means graphically

When a straight line intersects a cubic curve, how many intersection points can they normally have? Under what special situation can they share a point where the line just touches the cubic without crossing it?

Use the idea of a double root

If the line just touches (is tangent to) the cubic at some $x$ -value, that $x$ -value is a double root of the equation you got after eliminating $y$ . A double root is a number that makes both the function and its derivative equal to zero.

Set up and solve the system for the double root

Let $f(x)$ be the cubic you got after eliminating $y$ . Write equations for $f(r)=0$ and $f'(r)=0$ for the double root $r$ , substitute one into the other to find $r$ , and then use that to solve for $m$ .

Desmos Guide

Graph the cubic

In Desmos, enter the cubic as y = x^3 - 4x to see its graph.

Graph the line family with a slider

Type y = m x + 2. Desmos will prompt you to add a slider for m; create the slider so you can vary the slope of the line.

Adjust m to get exactly two intersection points

Move the m slider and watch how the line intersects the cubic. You are looking for the value of m where the line intersects the cubic in exactly two points, with one point where the line just touches the curve (tangent) instead of crossing it.

Read off and verify the m-value

When you see the line just touching the cubic at one point and crossing at another (two distinct intersection points), click the intersection points to confirm there are only two. Then read the corresponding value of m from the slider; that is the value that makes the system have exactly two distinct real solutions.

Step-by-step Explanation

Combine the equations into one in terms of x

The intersection points of the graphs occur where the $y$ -values are equal, so set the right-hand sides equal:

x^3 - 4x = mx + 2.

Move everything to one side:

x^3 - 4x - mx - 2 = 0

so the intersection $x$ -values satisfy

f(x) = x^3 - (m+4)x - 2 = 0.

The real solutions of $f(x)=0$ are exactly the $x$ -coordinates of the solutions to the system.

Interpret “exactly two distinct real solutions” for a cubic

A cubic with real coefficients can have:

three distinct real roots,
one real root and two complex roots, or
a repeated real root and one other real root.

To get exactly two distinct real solutions for the system, the cubic must have:

one real root that is a double root (the line is tangent to the cubic there), and
one other real root.

So $f(x)$ must have a double root. A double root $r$ is a number where $f(r)=0$ and the derivative $f'(r)=0$ .

Use the derivative to express the double-root condition

First find the derivative of $f(x)$ :

f(x) = x^3 - (m+4)x - 2,

f'(x) = 3x^2 - (m+4).

If $r$ is the double root, then

\begin{aligned} f(r) &= r^3 - (m+4)r - 2 = 0, \\ f'(r) &= 3r^2 - (m+4) = 0. \end{aligned}

From the second equation,

m+4 = 3r^2.

We will substitute this into the first equation to solve for $r$ .

Solve for the double root r

Substitute $m+4 = 3r^2$ into $f(r)=0$ :

\begin{aligned} 0 &= r^3 - (m+4)r - 2 \\ &= r^3 - (3r^2)r - 2 \\ &= r^3 - 3r^3 - 2 \\ &= -2r^3 - 2. \end{aligned}

-2r^3 - 2 = 0.

Add $2$ to both sides and then divide by $-2$ (or multiply both sides by $-1$ first):

2r^3 + 2 = 0 \quad\Rightarrow\quad r^3 = -1.

Thus the double root is

r = -1.

Find m and confirm the number of solutions

Use $r=-1$ in the relation $m+4 = 3r^2$ :

\begin{aligned} m+4 &= 3(-1)^2 \\ &= 3. \end{aligned}

m = 3 - 4 = -1.

With $m=-1$ , the equation for the intersections is

x^3 - (m+4)x - 2 = x^3 - 3x - 2.

This factors as

x^3 - 3x - 2 = (x+1)^2(x-2),

which has real roots $x=-1$ (double) and $x=2$ (simple). These give exactly two distinct real $x$ -values for the system, so the required value is

\boxed{-1}.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

Study Hints

Strategy

Hints

Desmos Guide

Step-by-step Explanation