Question 181·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

0 / 233

A system of equations is given by

y = 2x + 1 \qquad\text{and}\qquad y = (x - 3)^2.

One solution to the system is $(x, y)$ with $x < 3$ . What is the value of $x$ ?

A nonlinear system has equations that must hold at the same point, with at least one curved graph. Graph both equations in Desmos and click their intersections; each crossing gives a possible $(x,y)$ . Use the inequality on $x$ to choose a crossing, then evaluate an exact answer choice because Desmos labels intersections with rounded decimals.

Hints

Hint 1
A solution to a system makes both equations true at once, so it sits where their graphs cross. Enter each equation on its own Desmos line to see every possible solution.
Hint 2
In a point written $(x,y)$ , $x$ comes first. Compare the first coordinate of each crossing with $3$ . Then evaluate an exact answer choice to see whether it matches the coordinate Desmos shows.

Step-by-step

Approach 1: Graph the system and match an exact choice
Step 1
Graph both equations
Type $y=2x+1$ and $y=(x-3)^2$ on separate lines. Desmos shows a line and a parabola, a U-shaped graph, crossing twice. Each crossing is a solution because the point lies on both graphs.
Step 2
Select the crossing with the required x-coordinate
Click both intersections, the places where the graphs meet. Desmos labels them about $(1.172,3.343)$ and $(6.828,14.657)$ . The first number in each pair is $x$ , so only the left crossing has $x<3$ . Use the condition to choose the intersection before matching an exact answer.
Step 3
Match the coordinate to an exact expression
Type $4-2\sqrt{2}$ on a new line. Desmos gives about $1.172$ , matching the selected $x$ -coordinate. The intersection label is rounded, but the choice gives its exact value. So the requested $x$ is $4-2\sqrt{2}$ . Choice D.

Approach 2: Find the exact roots by completing the square
Step 1
Set the two expressions for y equal
At a solution, both equations give the same $y$ . Set their right sides equal: $(x-3)^2=2x+1$ .
Step 2
Expand the square
Expand $(x-3)^2$ so you can combine the $x$ terms: $x^2-6x+9=2x+1$ .
Step 3
Move the x term and constant
Subtract $2x+9$ from both sides: $x^2-8x=-8$ .
Step 4
Add the missing piece of a square
Half the coefficient $-8$ is $-4$ , and $(-4)^2=16$ . Add $16$ to both sides to make the left side a perfect square: $x^2-8x+16=-8+16$ .
Step 5
Rewrite as one square
Since $(x-4)^2=x^2-8x+16$ and $-8+16=8$ , rewrite the equation as one square: $(x-4)^2=8$ .
Step 6
Keep both square roots
Take square roots, keeping both signs because either a positive or a negative number can square to $8$ : $x-4=\pm\sqrt{8}$ .
Step 7
Isolate x
Add $4$ to both sides: $x=4\pm\sqrt{8}$ .
Step 8
Simplify the radical
Use $\sqrt{8}=\sqrt{4\cdot2}=2\sqrt{2}$ to simplify both roots: $x=4\pm2\sqrt{2}$ .
Step 9
Apply the condition
Because $2\sqrt{2}>1$ , the minus root satisfies $4-2\sqrt{2}<3$ ; the plus root is greater than $3$ . So the requested $x$ is $4-2\sqrt{2}$ . Choice D.

Question 181·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