Question 217·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations:

\begin{aligned} x + y &= 5 \\[1.4em] x^2 - y &= 11 \end{aligned}

What are the possible values of $x$ ?

The squared term signals a linear and nonlinear system: a line can meet a curve twice. Graph both equations in Desmos to locate their intersections, then add the equations to cancel a variable. Use the quadratic formula to match exact radical choices. Each intersection gives one possible $x$ -value, so keep both. Graph decimals alone aren't exact radical answers.

Hints

Hint 1
A solution to a system is a point that makes both equations true. Graph the equations as written and find both places they meet. Which number in each point is the $x$ -value?
Hint 2
Adding equations can eliminate a variable, meaning it cancels out. Here $+y$ and $-y$ cancel when added. What equation containing only $x$ remains?
Hint 3
The quadratic formula uses the coefficients from an equation equal to zero. Move every term to one side before identifying them; in particular, check the sign of the constant term.

Step-by-step

Graph both solutions, then find their exact forms
Step 1
Locate both intersections
Type $x+y=5$ and $x^2-y=11$ on separate Desmos lines. Click both intersections, the points that satisfy both equations. Desmos shows about $(-4.531,9.531)$ and $(3.531,1.469)$ . The first coordinate is $x$ , so you need two exact $x$ -values, not one rounded graph value.
Step 2
Cancel $y$ by adding
Add the two equations. This is elimination: the $+y$ and $-y$ cancel, leaving an equation in $x$ alone:
$\begin{aligned}(x+y)+(x^2-y)&=5+11\x^2+x&=16.\end{aligned}$
Step 3
Identify the signed coefficients
Subtract $16$ to write the quadratic, an equation containing $x^2$ , in $ax^2+bx+c=0$ form:
$x^2+x-16=0.$
So $a=1$ , $b=1$ , and $c=-16$ . The $16$ was positive on the right, but it changes sign when moved to the left.
Step 4
Find the number under the square root
The discriminant is $b^2-4ac$ , the part under the square root in the quadratic formula. Type $1^2-4(1)(-16)$ into Desmos. It prints $65$ ; subtracting a negative adds $64$ . Since $65$ is positive, the formula gives two real values.
Step 5
Use the formula for the exact values
For $ax^2+bx+c=0$ , the quadratic formula is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ . Substitute the signed coefficients and keep both signs of the square root:
$\begin{aligned}x&=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[1.4em]&=\frac{-1\pm\sqrt{65}}{2(1)}\\[1.4em]&=\frac{-1\pm\sqrt{65}}{2}.\end{aligned}$
These are the two possible values of $x$ ; $\pm$ gives the plus and minus versions. Choice D.

Question 217·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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