Question 101·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The system of equations is given by

\begin{cases} y = x^2 - 4x + 3 \\ y = 2x - 1 \end{cases}

The two graphs intersect at two points. What is the $x$ -coordinate of the point with the greater $x$ -value?

A nonlinear system with a line and a parabola can have two intersection points, each of which satisfies both equations. Graph the equations in Desmos and compare the requested coordinates of those points. If the choices contain square roots, evaluate a choice to match its exact form to the decimal Desmos shows. Don't stop at the first intersection you find.

Hints

Hint 1
An intersection is a point on both graphs, so its coordinates make both equations true. A line can cross a parabola twice. Which number in each point tells you whether it has the greater $x$ -value?
Hint 2
Desmos shows intersection coordinates as rounded decimals, even when a choice contains a square root. After you identify the right-hand point, evaluate a choice in Desmos to see whether its decimal matches that point's first coordinate.

Step-by-step

Approach 1: Graph the two intersections
Step 1
Find the left intersection
The squared $x$ makes the first graph a parabola, a U-shaped curve. Type both equations on separate Desmos lines. Click their left intersection; Desmos labels it approximately $(0.764,0.528)$ . That point satisfies both equations, but its first coordinate is the smaller $x$ -value, so don't stop there.
Step 2
Match the greater x-coordinate to an exact choice
Click the right intersection; Desmos labels it approximately $(5.236,9.472)$ . An ordered pair $(x,y)$ lists $x$ first, so compare $5.236$ with the choices. Type $3+\sqrt{5}$ on a new line; Desmos displays about $5.236$ , a match. Compare the first coordinates, not the heights of the points. The greater $x$ -coordinate is $3+\sqrt{5}$ . Choice B.

Approach 2: Get the exact coordinates by completing the square
Step 1
Set the two y-values equal
At an intersection, both equations give the same $y$ , so set their right sides equal:
$x^2-4x+3=2x-1.$
Step 2
Make a perfect square
Subtract $2x-1$ from both sides:
$x^2-6x+4=0.$
Subtract $4$ from both sides:
$x^2-6x=-4.$
For completing the square, halve $-6$ to get $-3$ , then square it to get $9$ . Add $9$ to both sides:
$x^2-6x+9=-4+9.$
Rewrite the left side as a square and the right side as $5$ :
$(x-3)^2=5.$
Step 3
Keep both square roots
Take the square root of each side, keeping both signs: squaring either $\sqrt{5}$ or $-\sqrt{5}$ gives $5$ .
$x-3=\pm\sqrt{5}.$
Step 4
Choose the greater x-value
Add $3$ to both sides:
$x=3\pm\sqrt{5}.$
Because $\sqrt{5}$ is positive, the plus sign gives the greater $x$ -coordinate, $3+\sqrt{5}$ . Choice B.

Question 101·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

Strategy

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Step-by-step

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Strategy

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