The graphs of the following equations intersect at exactly one point in the -plane. What is the value of at this point of intersection?

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SAT Nonlinear Equations & Systems Question 61 | Free SAT Prep | Aniko

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Question 61·Easy·Nonlinear Equations in One Variable; Systems in Two Variables

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The graphs of the following equations intersect at exactly one point $(x,y)$ in the $xy$ -plane.

y = \sqrt{x + 9}

y = x - 3

What is the value of $x$ at this point of intersection?

For intersection problems, set the right-hand sides of the equations equal because the $y$ -values are the same where the graphs meet. When a square root is involved, first write an inequality for any domain or sign restrictions (like making sure the square-root output is nonnegative), then square both sides to remove the root, solve the resulting equation, and finally substitute each solution back into the original equations to eliminate any extraneous solutions created by squaring.

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Use the idea of intersection

At a point where two graphs intersect, what can you say about their $y$ -values for the same $x$ ?

Form an equation with one variable

Set $\sqrt{x+9}$ equal to $x - 3$ and think about how to remove the square root so you can solve for $x$ .

Be careful after squaring

After you square both sides, you might get more than one solution for $x$ . How can you check whether each solution works in the original equations?

Think about the square root's properties

The expression $\sqrt{x+9}$ is always nonnegative. What does that tell you about what $x - 3$ must be?

Desmos Guide

Graph both equations

In Desmos, enter the two equations on separate lines:

y = sqrt(x + 9)
y = x - 3

Locate the intersection point

Adjust the zoom so that the curve and the line are both clearly visible where they cross. Then click or tap on their intersection point; Desmos will display its coordinates $(x,y)$ .

Read off the needed value

From the displayed intersection coordinates, note the x-coordinate; that x-value is the answer to the question.

Step-by-step Explanation

Set the equations equal at the point of intersection

At a point of intersection, the $y$ -values of the two graphs are equal.

Set the right-hand sides equal:

\sqrt{x+9} = x - 3

Also remember: because $\sqrt{x+9}$ is always nonnegative, the right side $x-3$ must be greater than or equal to 0, so $x \ge 3$ .

Eliminate the square root and solve the quadratic

Square both sides to get rid of the square root:

x+9 = (x-3)^2

Now expand and simplify:

\begin{aligned} x+9 &= x^2 - 6x + 9 \\ 0 &= x^2 - 7x \end{aligned}

Factor the quadratic:

0 = x(x - 7)

So the possible solutions are $x = 0$ or $x = 7$ .

Check which solution actually works in the original equations

We must check each possible $x$ in the original equation $\sqrt{x+9} = x - 3$ .

For $x = 0$ :
- Left side: $\sqrt{0 + 9} = \sqrt{9} = 3$ .
- Right side: $0 - 3 = -3$ .
- $3 \ne -3$ , so $x = 0$ is an extraneous solution (introduced by squaring).
For $x = 7$ :
- Left side: $\sqrt{7 + 9} = \sqrt{16} = 4$ .
- Right side: $7 - 3 = 4$ .
- $4 = 4$ , so $x = 7$ does satisfy the original equation.

Therefore, the $x$ -value at the point of intersection is $\boxed{7}$ .

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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