Question 128·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

The system of equations is given by

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

If $(x, y)$ is a solution to the system above, which of the following could be the value of $x$ ?

A nonlinear system with a squared $x$ pairs a curve with a line, so it may have two solutions. Graph both equations in Desmos and click every intersection; the first coordinate gives $x$ . If a choice contains a square root, evaluate it as a decimal to compare with the graph. Don't round an intersection to a nearby integer.

Hints

Hint 1
In a system, a solution must satisfy both equations. On a graph, that happens wherever the two graphs cross. What would you type to see those crossings?
Hint 2
An ordered pair $(x,y)$ lists $x$ first and $y$ second. Click each crossing and read its first coordinate. A line and a parabola can cross twice, so check both points.
Hint 3
Desmos labels intersections with rounded decimals, while a radical contains a square root. Evaluate exact expressions as decimals before comparing them with the graph; rounding an intersection to an integer can give you a non-solution.

Step-by-step

Approach 1: Graph the system and match a choice
Step 1
Find both intersections
Type $y=x^2-4x$ and $y=2x-4$ on separate Desmos lines. Click both intersections: Desmos shows approximately $(0.764,-2.472)$ and $(5.236,6.472)$ . An intersection lies on both graphs, so it satisfies both equations. The first coordinate is $x$ ; $0.764$ isn't $1$ .
Step 2
Match an exact choice to an x-coordinate
The graph gives rounded decimals, but a choice is written exactly. Type $3+\sqrt{5}$ on a new line. Desmos prints approximately $5.236$ , matching the right-hand intersection's $x$ -coordinate. So $3+\sqrt{5}$ could be the value of $x$ . Choice C.

Approach 2: Find the exact values by completing the square
Step 1
Set the two y-values equal
At a solution, both equations give the same $y$ , so their right sides must be equal. This removes $y$ :
$x^2-4x=2x-4.$
Step 2
Collect the x-terms
Subtract $2x$ from both sides: $x^2-6x=-4$ . Keep the $-4$ ; dropping it changes the solutions.
Step 3
Add the missing square term
To make a perfect square, halve $-6$ to get $-3$ , then square it to get $9$ . Add $9$ to both sides:
$x^2-6x+9=-4+9.$
Step 4
Write the completed square
The left side is $(x-3)^2$ because squaring $x-3$ produces $x^2-6x+9$ . Rewrite it and calculate the right side:
$(x-3)^2=5.$
Step 5
Keep both square roots
Both $\sqrt{5}$ and $-\sqrt{5}$ square to $5$ . Take the square root with both signs:
$x-3=\pm\sqrt{5}.$
Step 6
Identify a possible x-value
Add $3$ to both sides: $x=3\pm\sqrt{5}$ . The plus-sign value, $3+\sqrt{5}$ , is one possible value of $x$ for the system. Choice C.

Question 128·Medium·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