Question 907·Medium·Math

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Consider the system of equations

\begin{cases} y = 3x - 2 \\ y = x^2 - 5 \end{cases}

What is the positive solution for $x$ that satisfies the system?

A linear-quadratic system pairs a line with a parabola, so the graphs may meet more than once. Graph both equations in Desmos and click their intersections. Each intersection solves both equations, but the question may want only one coordinate. Check the sign of the requested coordinate, then evaluate an exact answer choice to match Desmos’s rounded value.

Hints

Hint 1
A solution to a system is one point that makes both equations true. On a graph, look where the line and parabola cross each other, not where either graph crosses an axis.
Hint 2
At an intersection, the first coordinate is $x$ and the second is $y$ . The word positive tells you which $x$ -coordinate to keep when you click the crossing points.
Hint 3
Desmos labels a crossing with rounded decimals, while the choices contain radicals, or square roots. Evaluate an exact expression on a new line and compare its decimal with the positive $x$ -coordinate.

Step-by-step

Approach 1: Graph the intersections and match a choice

Step 1
Find the positive intersection
Type $y=3x-2$ and $y=x^2-5$ on separate Desmos lines. An intersection lies on both graphs, so it satisfies both equations. Click the crossings: their $x$ -coordinates are about $-0.791$ and $3.791$ . Keep the positive $x$ -coordinate, about $3.791$ .
Step 2
Match the rounded coordinate to an exact choice
Type $\frac{3+\sqrt{21}}{2}$ on a new line. Desmos prints about $3.791$ , matching the positive intersection’s $x$ -coordinate. The graph gives a rounded decimal; this radical gives the exact positive solution for $x$ . Choice B.

Approach 2: Get the exact roots with the quadratic formula

Step 1
Set the two values of y equal
At a solution, both equations give the same $y$ . Set their right sides equal:
$x^2-5=3x-2.$
Step 2
Move the constant
Add $5$ to both sides:
$x^2=3x+3.$
Step 3
Put the quadratic in standard form
Subtract $3x+3$ from both sides so one side is $0$ :
$x^2-3x-3=0.$
Step 4
Substitute into the quadratic formula
For a quadratic $ax^2+bx+c=0$ , the formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . Here $a=1$ , $b=-3$ , and $c=-3$ . Keep the negative signs in parentheses as you substitute:
$x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-3)}}{2(1)}.$
Step 5
Simplify the signed numbers
Square $-3$ and simplify $-4(1)(-3)$ , which is positive:
$x=\frac{3\pm\sqrt{9+12}}{2}.$
Step 6
Finish the exact roots
Add the numbers under the square root:
$x=\frac{3\pm\sqrt{21}}{2}.$
Step 7
Keep the positive root
Since $\sqrt{21}>3$ , subtracting it gives a negative numerator. So the plus sign gives the positive solution for $x$ , $\frac{3+\sqrt{21}}{2}$ . Choice B.

Question 907·Medium·Math

Strategy

Hints

Step-by-step

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Strategy

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Step-by-step

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