What is the product of all possible values of that satisfy the system of equations above?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 191 | Free SAT Prep | Aniko

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Question 191·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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

What is the product of all possible values of $x$ that satisfy the system of equations above?

Your answer

(Express the answer as an integer)

For a system with one linear equation and one quadratic equation, substitution is usually the fastest method: isolate one variable (here, $y$ is already isolated), substitute into the other equation, and simplify to get a single quadratic in $x$ . Once you have a quadratic, factor if possible (it’s usually chosen to factor nicely on the SAT) to get the two $x$ -values. When the problem asks for something like the product of the solutions, you can either multiply the two values directly or, if you’re comfortable with Vieta’s formulas, use the fact that for $ax^2+bx+c=0$ , the product of the solutions is $c/a$ to save time.

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Use substitution

Since the first equation already has $y$ by itself, try substituting $y = x^2 - 4$ into the second equation $x + y = 2$ .

Form a quadratic in $x$

After you substitute for $y$ , combine like terms and move everything to one side so the equation looks like $x^2 + bx + c = 0$ .

Factor and then find the product

Once you have the quadratic, factor it (or use the quadratic formula) to find both $x$ -values. Then multiply those two values together, as the question asks for their product.

Desmos Guide

Enter the two equations

Type y = x^2 - 4 on one line and y = 2 - x on another line in Desmos. These represent the two equations in the system.

Find the intersection points

Look at where the parabola and the line intersect. Click (or tap) each intersection point to see its coordinates, and note the two $x$ -values from those points.

Compute the product in Desmos

In a new expression line, type the product of the two $x$ -values you noted (for example, (first x-value) * (second x-value)). The resulting output is the product of all possible $x$ -values that satisfy the system.

Step-by-step Explanation

Substitute to get one equation in $x$

From the system,

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

$y$ is already isolated in the first equation. Substitute $y = x^2 - 4$ into the second equation:

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

Now you have an equation involving only $x$ .

Simplify to a quadratic equation

Combine like terms and move everything to one side:

\begin{aligned} x + x^2 - 4 &= 2 \\ x^2 + x - 4 &= 2 \\ x^2 + x - 4 - 2 &= 0 \\ x^2 + x - 6 &= 0. \end{aligned}

So the $x$ -values that satisfy the system must satisfy the quadratic equation $x^2 + x - 6 = 0$ .

Factor and find all possible $x$ -values

Factor the quadratic:

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

Set each factor equal to zero:

$x + 3 = 0$ gives $x = -3$ .
$x - 2 = 0$ gives $x = 2$ .

So the system has two possible $x$ -values: $-3$ and $2$ .

Compute the product of the $x$ -values

The question asks for the product of all possible values of $x$ .

Multiply the two values you found:

(-3) \cdot 2 = -6.

Therefore, the product of all possible $x$ -values that satisfy the system is $-6$ .

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