Question 54·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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y = x^2 - 4x - 1 \\[1.4em] y = 2x + 5

The ordered pair $(x, y)$ satisfies the system of equations above. What is the product of all possible values of $y$ ?

A nonlinear system pairs a curve with another graph, such as a line. Their intersections are the ordered pairs that satisfy both equations. Graph both equations in Desmos and click every intersection, then multiply the coordinates the question asks for. The tempting slip is to multiply the first coordinates when the question wants the second ones.

Hints

Hint 1
A solution to a system makes both equations true. On a graph, that means a point where the curve and line cross. Find every crossing before deciding which numbers to multiply.
Hint 2
In an ordered pair $(x,y)$ , the second coordinate is $y$ . Click each intersection and record its second coordinate, not its first. What does the question ask you to do with those two values?
Hint 3
Desmos may show rounded coordinates when you click a point. Multiply the two $y$ -coordinates, then check whether a small difference from a whole-number choice comes from that rounding.

Step-by-step

Approach 1: Graph both equations and multiply the y-coordinates
Step 1
Find every solution on the graph
Type $y=x^2-4x-1$ and $y=2x+5$ on separate Desmos lines. A solution to the system lies on both graphs, so click both intersections. Desmos shows about $(-0.873,3.254)$ and $(6.873,18.746)$ . The second coordinate at each intersection is a possible $y$ -value.
Step 2
Multiply the requested values
Type $3.254\cdot18.746$ on a new line. Desmos shows $60.999484$ , slightly below $61$ because the clicked coordinates were rounded. You're multiplying the two possible $y$ -values, not the $x$ -values, so their product is $61$ . Choice C.

Approach 2: Get the exact product from the quadratic
Step 1
Make the two expressions for y equal
At a solution, both equations give the same $y$ , so set their right sides equal: $x^2-4x-1=2x+5$ .
Step 2
Write one quadratic equation
Subtract $2x+5$ from both sides to get one equation for $x$ : $x^2-6x-6=0$ .
Step 3
Find the roots' sum and product
Call the two solutions for $x$ $r$ and $s$ . A root is a value that makes a quadratic equal $0$ . Expanding its factors shows why the coefficients give the roots' sum and product: $a(x-r)(x-s)=ax^2-a(r+s)x+ars$ . In $ax^2+bx+c=0$ , this means $r+s=-\frac{b}{a}$ and $rs=\frac{c}{a}$ . Here $a=1$ , $b=-6$ , and $c=-6$ , so you don't need either root separately: $r+s=-\frac{-6}{1}=6,\qquad rs=\frac{-6}{1}=-6$ .
Step 4
Write the two possible y-values
The line says $y=2x+5$ . Put each $x$ -solution into it: $y_1=2r+5,\qquad y_2=2s+5$ . These are both possible $y$ -values, not the $x$ -values whose sum and product you found.
Step 5
Multiply the y-expressions
Multiply every term in the two expressions: $y_1y_2=(2r+5)(2s+5)=4rs+10r+10s+25$ .
Step 6
Group the terms you know
Group $10r+10s$ so you can use the known sum $r+s$ : $y_1y_2=4rs+10(r+s)+25$ .
Step 7
Calculate the exact product
Type $4(-6)+10(6)+25$ into Desmos, using $rs=-6$ and $r+s=6$ . It shows $61$ , the exact product of all possible $y$ -values. Choice C.

Question 54·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

Hints

Step-by-step

Lessons that teach this

Study Hints

Strategy

Hints

Step-by-step

Lessons that teach this