In the -plane, consider the system of equations where is a real constant. The system has exactly one real solution. What is the sum of all possible values of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 138 | Free SAT Prep | Aniko

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Question 138·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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In the $xy$ -plane, consider the system of equations

\begin{cases} x^2 + y^2 = 20 \\[4pt] y = \dfrac{1}{2}x + b \end{cases}

where $b$ is a real constant. The system has exactly one real solution. What is the sum of all possible values of $b$ ?

For circle–line system questions asking for “exactly one solution,” think: tangent line. The fastest algebraic approach is to substitute the line equation into the circle equation to get a quadratic in one variable, then use the discriminant condition $B^2 - 4AC = 0$ to enforce exactly one real intersection. Solve for the parameter (here, $b$ ), list all resulting values, and carefully do whatever the question asks with them (such as summing them), watching out not to overlook symmetric positive/negative pairs.

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Think about the graph

One equation is a circle and the other is a line. What must be true about how the line meets the circle if there is exactly one point that satisfies both equations?

Reduce the system to one variable

Use the expression for $y$ from the line, $y = \dfrac{1}{2}x + b$ , and substitute it into the circle equation $x^2 + y^2 = 20$ to get an equation in terms of $x$ and $b$ only.

Use the quadratic discriminant

After substitution, you will get a quadratic equation in $x$ . For there to be exactly one real solution for $x$ , what must the discriminant $B^2 - 4AC$ of that quadratic be?

Finish with the parameter b

Once you set the discriminant to the needed value and solve for $b$ , remember the question is asking for the sum of all possible values of $b$ , not just one of them.

Desmos Guide

Graph the circle

In Desmos, type x^2 + y^2 = 20 to graph the circle centered at the origin with radius $\sqrt{20}$ .

Graph the family of lines with parameter b

Type y = (1/2)x + b. Desmos will prompt you to create a slider for b; create the slider so you can move the line up and down.

Find b-values where the line is tangent

Move the b slider and watch how the line intersects the circle. Look for the two positions where the line just touches the circle at exactly one point (tangent) instead of cutting through it. Note the two corresponding values of b from the slider.

Compute the requested sum

Take the two b values you observed in Desmos when the line was tangent to the circle and add them together to get the sum requested in the problem.

Step-by-step Explanation

Interpret what “exactly one real solution” means

The equations

\begin{cases} x^2 + y^2 = 20 \\ y = \dfrac{1}{2}x + b \end{cases}

describe a circle and a line in the $xy$ -plane.

The circle $x^2 + y^2 = 20$ is centered at $(0,0)$ .
The line $y = \dfrac{1}{2}x + b$ has slope $\dfrac{1}{2}$ and $y$ -intercept $b$ .

“Exactly one real solution” means the line touches the circle at exactly one point, so the line is tangent to the circle.

Substitute the line equation into the circle equation

Substitute $y = \dfrac{1}{2}x + b$ into $x^2 + y^2 = 20$ :

x^2 + \left(\frac{1}{2}x + b\right)^2 = 20.

Expand the square:

\left(\frac{1}{2}x + b\right)^2 = \frac{1}{4}x^2 + bx + b^2.

So the equation becomes

x^2 + \frac{1}{4}x^2 + bx + b^2 = 20.

Combine like terms:

\frac{5}{4}x^2 + bx + b^2 - 20 = 0.

Multiply by $4$ to clear the fraction:

5x^2 + 4bx + 4b^2 - 80 = 0.

This is a quadratic equation in $x$ .

Use the discriminant condition for exactly one real solution

For a quadratic in $x$ of the form $Ax^2 + Bx + C = 0$ , there is exactly one real solution when the discriminant $B^2 - 4AC$ is $0$ .

Here,

$A = 5$ ,
$B = 4b$ ,
$C = 4b^2 - 80$ .

Set the discriminant equal to $0$ :

(4b)^2 - 4(5)(4b^2 - 80) = 0.

Simplify:

16b^2 - 20(4b^2 - 80) = 0 \\[0.7em] 16b^2 - 80b^2 + 1600 = 0 \\[0.7em] -64b^2 + 1600 = 0.

Solve for $b^2$ :

64b^2 = 1600 \\[0.7em] b^2 = 25.

So the possible values of $b$ are $b = 5$ and $b = -5$ .

Find the requested sum of all possible values of b

We have found two possible values of $b$ : $5$ and $-5$ .

The problem asks for the sum of all possible values of $b$ :

5 + (-5) = 0.

So, the correct answer is 0.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation