SAT Nonlinear Equations & Systems Question 117 | Free SAT Prep | Aniko

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Question 117·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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In the $xy$ -plane, how many points satisfy the system of equations below?

\begin{cases} y = x^{4} - 4x^{2} + 3 \\ y = x + 1 \end{cases}

For systems where a polynomial equals a line, think "intersections": set the two expressions equal to form a single equation in $x$ , simplify to 0, then solve that equation. Start by looking for easy integer roots to factor the polynomial, reduce the degree step by step (using synthetic/long division), and use the quadratic formula for any remaining quadratic factor. Finally, count the distinct real $x$ -values you find, since each corresponds to one solution $(x,y)$ via the simpler equation (usually the line).

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Think in terms of intersections

A point that satisfies both equations lies on both graphs. How do you usually find where two graphs intersect?

Set the right-hand sides equal

Since both expressions equal $y$ , set $x^4 - 4x^2 + 3$ equal to $x + 1$ and rearrange to form one equation equal to 0.

Factor to find $x$ -values

Once you have the quartic equation in $x$ , try to factor it by finding simple integer roots like $-2,-1,0,1,2$ , then use division to reduce the degree.

Use the quadratic formula if needed

If part of the expression factors to a quadratic that you cannot factor easily, apply the quadratic formula to find its roots.

Desmos Guide

Enter the two functions

In one expression line, type y = x^4 - 4x^2 + 3. In another line, type y = x + 1 to graph both equations on the same axes.

Adjust the view if needed

Use zoom out (−) and drag the graph so you can see where the quartic curve and the line cross each other over a reasonable range of $x$ (for example, from about $x = -3$ to $x = 3$ ).

Find and count intersections

Click on each point where the line and the quartic curve intersect; Desmos will display their coordinates. Count how many distinct intersection points you see—that count is the number of solutions to the system.

Step-by-step Explanation

Set the equations equal and form one equation in $x$

The solutions are the intersection points of the two graphs, so their $x$ - and $y$ -coordinates must satisfy both equations.

Set the right sides equal:

x^4 - 4x^2 + 3 = x + 1.

Move everything to one side:

x^4 - 4x^2 + 3 - x - 1 = 0 \\[0.7em] x^4 - 4x^2 - x + 2 = 0.

Now we just need to solve this quartic (degree 4) equation for $x$ .

Factor the quartic equation

Try to factor $x^4 - 4x^2 - x + 2$ by finding simple roots (values of $x$ that make it zero).

Test small integers:

For $x = -1$ :

(-1)^4 - 4(-1)^2 - (-1) + 2 = 1 - 4 + 1 + 2 = 0,

so $x = -1$ is a root, and $(x + 1)$ is a factor.

Divide $x^4 - 4x^2 - x + 2$ by $(x+1)$ (using synthetic or long division) to get:

(x + 1)(x^3 - x^2 - 3x + 2).

Now factor the cubic $x^3 - x^2 - 3x + 2$ .

Test $x = 2$ :

2^3 - 2^2 - 3(2) + 2 = 8 - 4 - 6 + 2 = 0,

so $x = 2$ is a root, and $(x - 2)$ is a factor. Divide to get:

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

So the full factorization is

x^4 - 4x^2 - x + 2 = (x + 1)(x - 2)(x^2 + x - 1).

Solve for all $x$ -values and count the solutions

Set each factor equal to zero:

$x + 1 = 0 \Rightarrow x = -1$
$x - 2 = 0 \Rightarrow x = 2$
$x^2 + x - 1 = 0$ .

Solve $x^2 + x - 1 = 0$ using the quadratic formula:

x = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{-1 \pm \sqrt{5}}{2}.

So the four distinct real $x$ -values where the graphs intersect are

x = -1,\; 2,\; \frac{-1 + \sqrt{5}}{2},\; \frac{-1 - \sqrt{5}}{2}.

Each $x$ gives one point $(x,y)$ on both graphs, with $y = x + 1$ . Therefore, there are exactly 4 solutions to the system.

Strategy

Hints

Desmos Guide

Step-by-step Explanation

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Strategy

Hints

Desmos Guide

Step-by-step Explanation