Question 113·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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In the $xy$ -plane, the graphs of $y = x^2 + 2x - 1$ and $y = -x + 5$ intersect at two points. What is the product of the $x$ -coordinates of these points?

Your answer

When two graphs intersect, set their right-hand sides equal to form an equation in one variable. If this gives a quadratic and the question asks for the sum or product of the intersection $x$ -coordinates, rewrite the equation in $ax^2+bx+c=0$ form and use the coefficient relationships: the sum of the roots is $-b/a$ , and the product of the roots is $c/a$ . This avoids unnecessary quadratic-formula work when only a combination of the solutions is needed.

Hints

Hint 1
Connect intersection points to equations
At intersection points, the two graphs have the same $x$ and $y$ values. How can you use this fact to write an equation involving only $x$ ?
Hint 2
Form and recognize the quadratic
After you set $x^2 + 2x - 1$ equal to $-x + 5$ and move all terms to one side, you should get a quadratic equation. Write it clearly in the form $ax^2 + bx + c = 0$ and identify $a$ , $b$ , and $c$ .
Hint 3
Use a shortcut for the product of solutions
Instead of solving the quadratic completely, recall: for $ax^2 + bx + c = 0$ , how can you find the product of the two solutions using only $a$ and $c$ ?
Hint 4
Apply the shortcut
Use your identified values of $a$ and $c$ from the quadratic you formed to compute $c/a$ . That value is the product of the $x$ -coordinates of the intersection points.

Desmos guide

Step 1
Graph the equations
Algebra is faster for this question, but Desmos can verify the intersections. Enter $y=x^2+2x-1$ and $y=-x+5$ on separate expression lines.
Step 2
Inspect the intersection points
Click each point where the parabola and line intersect. Desmos displays the two intersection coordinates and confirms that there are two corresponding $x$ -values.
Step 3
Verify the exact product
For an exact calculator check, enter
$\left(\frac{-3-\sqrt{33}}{2}\right)\left(\frac{-3+\sqrt{33}}{2}\right).$
This multiplies the two solutions of $x^2+3x-6=0$ and verifies the algebraic product.

Step-by-step

Step 1
Set the equations equal to find the intersection $x$ -values
At intersection points, the $y$ -values of the two graphs are equal, so set the right sides equal:
$x^2 + 2x - 1 = -x + 5$
Move all terms to one side to get a quadratic equation:
$\begin{aligned} x^2 + 2x - 1 + x - 5 &= 0 \\[1.4em] x^2 + 3x - 6 &= 0. \end{aligned}$
The $x$ -coordinates of the intersection points are the solutions of $x^2+3x-6=0$ .
Step 2
Use the product relationship for a quadratic
You are asked for the product of the two $x$ -coordinates, not the coordinates themselves.

For a quadratic of the form
$ax^2+bx+c=0,$
with solutions $r_1$ and $r_2$ , the product of the solutions is
$r_1r_2=\frac{c}{a}.$
Therefore, the product can be found directly from the leading coefficient and constant term.
3
Step 3
Identify the coefficients and compute the product
Compare $x^2+3x-6=0$ with $ax^2+bx+c=0$ :

$a=1$

$b=3$

$c=-6$

The product of the two roots, which are the $x$ -coordinates of the intersection points, is
$\frac{c}{a}=\frac{-6}{1}=-6.$
So, the product of the $x$ -coordinates of the intersection points is $-6$ .

Question 113·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Desmos guide

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Desmos guide

Step-by-step

Lessons that teach this