SAT Nonlinear Equations & Systems Question 147 | Free SAT Prep | Aniko

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Question 147·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Consider the system of equations

\begin{cases} x^2 + y^2 = 8 \\[4pt] y = x + t \end{cases}

where $t$ is a real constant. If the system has exactly one real solution $(x, y)$ , which of the following could be the value of $t$ ?

When a system involves a circle and a line with a parameter and asks for 'exactly one real solution,' think: tangent line. The fastest algebraic method is to substitute the line into the circle to get a quadratic in one variable, then use the discriminant condition $b^2 - 4ac = 0$ for exactly one real solution and solve for the parameter. Always check your resulting parameter values against the answer choices, and remember that a positive discriminant means two solutions, zero means one, and negative means none.

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Relate the system to graphs

Think of $x^2 + y^2 = 8$ as a circle and $y = x + t$ as a straight line. What must be true about how they intersect if there is exactly one real solution?

Eliminate one variable

Use the equation $y = x + t$ and substitute it into $x^2 + y^2 = 8$ so that you get an equation in terms of $x$ and $t$ only.

Use quadratic properties

After substitution, you will have a quadratic equation in $x$ . Recall that a quadratic $ax^2 + bx + c = 0$ has exactly one real solution when $b^2 - 4ac = 0$ . Compute this discriminant in terms of $t$ .

Solve for t and compare to the options

Set the discriminant equal to 0, solve that equation for $t$ , and then see which of the answer choices matches one of the values you found.

Desmos Guide

Graph the circle

In Desmos, enter the equation x^2 + y^2 = 8 to draw the circle centered at the origin.

Test each answer choice as the line

For each option, type the corresponding line:

For $t = -5$ : enter y = x - 5.
For $t = -4$ : enter y = x - 4.
For $t = 2$ : enter y = x + 2.
For $t = 3$ : enter y = x + 3. Observe how many intersection points each line has with the circle.

Identify the correct t from the graph

Look for the line that just touches the circle at exactly one point (is tangent to it). The value of $t$ that produces this tangent line is the correct answer.

Step-by-step Explanation

Understand what 'exactly one real solution' means

The system

\begin{cases} x^2 + y^2 = 8 \\ y = x + t \end{cases}

represents a circle ( $x^2 + y^2 = 8$ ) and a line ( $y = x + t$ ).

Having exactly one real solution $(x,y)$ means the line intersects the circle at exactly one point. Algebraically, when we substitute the line into the circle, we will get a quadratic equation in $x$ that must have exactly one real solution.

Substitute to get a quadratic in x

From the second equation, $y = x + t$ . Substitute this into the circle equation:

Start with $x^2 + y^2 = 8$ .
Replace $y$ with $x+t$ :

x^2 + (x + t)^2 = 8.

Now expand $(x + t)^2$ :

x^2 + x^2 + 2tx + t^2 = 8.

Combine like terms:

2x^2 + 2tx + t^2 = 8.

Move 8 to the left to get a standard quadratic in $x$ :

2x^2 + 2tx + t^2 - 8 = 0.

This is a quadratic equation in $x$ with

$a = 2$ ,
$b = 2t$ ,
$c = t^2 - 8$ .

Use the discriminant condition for exactly one solution

For a quadratic $ax^2 + bx + c = 0$ :

It has exactly one real solution when the discriminant $b^2 - 4ac = 0$ .

Here, $a = 2$ , $b = 2t$ , and $c = t^2 - 8$ .

Compute the discriminant:

\begin{aligned} \Delta &= b^2 - 4ac \\ &= (2t)^2 - 4(2)(t^2 - 8) \\ &= 4t^2 - 8(t^2 - 8) \\ &= 4t^2 - 8t^2 + 64 \\ &= -4t^2 + 64. \end{aligned}

For exactly one real solution, set this equal to 0:

-4t^2 + 64 = 0.

Now solve this equation for $t$ .

Solve for t and match with the choices

Solve

-4t^2 + 64 = 0.

Add $4t^2$ to both sides:

64 = 4t^2.

Divide by 4:

16 = t^2.

t = \pm 4.

This means $t$ could be $4$ or $-4$ . Among the answer choices $-5, -4, 2, 3$ , the only one that matches is $-4$ , so that is the correct answer.

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