Question 81·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The ordered pair $(x,k)$ satisfies the system of equations

\begin{cases} k - x = 4 \\[4pt] x^2 + kx = 12 \end{cases}

What is the sum of all possible values of $k$ ?

Your answer

A linear-nonlinear system pairs a line with an equation containing a square or a product, so it may have more than one solution. Graph both equations in Desmos to find every intersection, then substitute the linear equation into the other one. If the question asks for a sum of possible values, the resulting quadratic’s middle coefficient can give that sum without finding each value exactly. Rounded graph labels alone don’t prove an exact answer.

Hints

Hint 1
An intersection is a point that satisfies both equations. Desmos uses $(x,y)$ for graph coordinates, so let $y$ stand for $k$ and click both intersections. Which coordinate of each point gives a possible $k$ ?
Hint 2
Use substitution to make an equation in $k$ alone. Rearrange the linear equation to get $x=k-4$ , then replace every $x$ in the other equation with $k-4$ .
Hint 3
A root is a value that makes an equation equal $0$ . For $ak^2+bk+c=0$ , the roots add to $-b/a$ . Once your equation is in that form, look at the coefficient of $k$ , not the constant term.

Step-by-step

Graph both solutions, then find their exact sum
Step 1
Find both solutions on the graph
Let $y$ stand for $k$ so Desmos can graph the system. Type $y-x=4$ and $x^2+xy=12$ , then click both intersections, the points that satisfy both equations. Desmos shows about $(-3.646,0.354)$ and $(1.646,5.646)$ . The second coordinate gives $k$ , so keep both points.
Step 2
Write x in terms of k
The graph labels are rounded, so use the linear equation to find an exact sum. Add $x$ to both sides: $k=x+4$ . Subtract $4$ from both sides:
$x=k-4.$
Step 3
Make one equation in k
Use substitution: each solution also satisfies $x^2+kx=12$ , so replace both occurrences of $x$ with $k-4$ :
$(k-4)^2+k(k-4)=12.$
Step 4
Expand and combine
Expand the square: $k^2-8k+16+k(k-4)=12$ . Distribute $k$ : $k^2-8k+16+k^2-4k=12$ . Combine like terms:
$2k^2-12k+16=12.$
Step 5
Put the quadratic equal to zero
Subtract $12$ from both sides: $2k^2-12k+4=0$ . Divide every term by $2$ : $k^2-6k+2=0$ . These are the possible $k$ -values, written as roots of a quadratic, an equation with a squared variable.
Step 6
Use the middle coefficient to get the sum
Call the two roots $k_1$ and $k_2$ . Because the coefficient of $k^2$ is $1$ , their factors expand as $(k-k_1)(k-k_2)=k^2-(k_1+k_2)k+k_1k_2$ . Match the coefficient of $k$ to the $-6k$ in your equation: $-(k_1+k_2)=-6$ . Multiply by $-1$ : $k_1+k_2=6$ . The middle coefficient gives the sum without finding either root separately. The sum of all possible values of $k$ is $6$ . Grid in 6.

Question 81·Hard·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