SAT Nonlinear Equations & Systems Question 120 | Free SAT Prep

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Question 120·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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Let $t$ be a real number. The equation

x^4 - 2tx^2 + t^2 - 9 = 0

has exactly two distinct real solutions. Which of the following values of $t$ satisfies this condition?

When you see a quartic that only involves $x^4$ , $x^2$ , and constants, immediately treat it as a quadratic in $x^2$ by substituting $y = x^2$ . Factor or solve that quadratic in $y$ , then translate back to $x$ and use sign reasoning: each positive $y$ gives two real $x$ , $y=0$ gives one, and negative $y$ gives none. On parameter questions, express these $y$ -values in terms of the parameter and use inequalities (like requiring one positive and one negative) to get a condition on the parameter, then simply match that condition against the answer choices instead of plugging in blindly.

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Simplify the type of equation

Notice the equation involves $x^4$ and $x^2$ only. How could you rewrite it as a quadratic equation in a new variable like $y = x^2$ ?

Factor in terms of the new variable

Once you have the quadratic in $y$ , try to factor it. Look for a perfect-square pattern like $(y - t)^2$ and then solve for $y$ .

Connect $y$ -solutions back to $x$ -solutions

Remember that $y = x^2$ . For each value of $y$ , think about when $x^2 = y$ gives 0, 1, or 2 real solutions. How many real $x$ come from a positive $y$ ? From $y = 0$ ? From a negative $y$ ?

Use sign analysis on the expressions involving $t$

You will get two expressions for $x^2$ , involving $t+3$ and $t-3$ . Use the signs of these expressions (positive, zero, or negative) to figure out when the total number of real $x$ -solutions is exactly 2, then see which choice fits that condition.

Desmos Guide

Use the factored form for clarity

In Desmos, type the expression in factored form using a parameter $t$ :

(x^2 - (t+3)) (x^2 - (t-3))

Then press Enter so Desmos graphs $y = (x^2 - (t+3))(x^2 - (t-3))$ .

Create a slider for $t$ or test each choice

Either create a slider for $t$ (Desmos will prompt you) or, for each answer choice, manually type that value in place of $t$ (for example, $t=-6$ , then $t=-2$ , etc.). After each substitution, look at how many times the graph crosses the $x$ -axis; each crossing corresponds to a real solution.

Identify the correct choice

Compare the graphs for $t=-6$ , $t=-2$ , $t=3$ , and $t=8$ . The correct value of $t$ is the one for which the graph intersects the $x$ -axis at exactly two distinct points, indicating exactly two real solutions for $x$ .

Step-by-step Explanation

Treat the equation as quadratic in $x^2$

The equation

x^4 - 2tx^2 + t^2 - 9 = 0

looks complicated, but notice it only has powers $x^4$ , $x^2$ , and a constant. Let $y = x^2$ . Then the equation becomes

y^2 - 2ty + t^2 - 9 = 0.

Now we have a quadratic equation in $y$ instead of a quartic in $x$ .

Factor the quadratic in $y$ and solve for $y$

Factor the quadratic in $y$ :

\begin{aligned} y^2 - 2ty + t^2 - 9 &= (y^2 - 2ty + t^2) - 9 \\ &= (y - t)^2 - 9. \end{aligned}

Set this equal to 0:

(y - t)^2 - 9 = 0 \implies (y - t)^2 = 9.

y - t = 3 \quad \text{or} \quad y - t = -3,

which gives

y = t + 3 \quad \text{or} \quad y = t - 3.

Recall that $y = x^2$ , so these correspond to $x^2 = t + 3$ and $x^2 = t - 3$ .

Relate $x^2$ values to number of real $x$ -solutions

For each equation $x^2 = k$ :

If $k > 0$ , there are two real solutions: $x = \pm\sqrt{k}$ .
If $k = 0$ , there is one real solution: $x=0$ .
If $k < 0$ , there are no real solutions.

Here the two possible $x^2$ values are $t+3$ and $t-3$ .

If both are positive, we get 4 real solutions.
If both are negative, we get 0 real solutions.
If one is positive and the other is negative, we get exactly 2 real solutions.
If one is zero and the other is positive, we get 3 real solutions.
If one is zero and the other is negative, we get 1 real solution.

We want exactly two distinct real solutions, so we need one of $t+3$ , $t-3$ to be positive and the other to be negative.

Find the condition on $t$ for one positive and one negative $x^2$ value

"One positive and one negative" means $(t+3)$ and $(t-3)$ have opposite signs, so their product must be negative:

(t+3)(t-3) < 0.

Compute the product:

(t+3)(t-3) = t^2 - 9.

So we need

t^2 - 9 < 0.

Solve this inequality:

t^2 < 9 \implies -3 < t < 3.

So any $t$ strictly between $-3$ and $3$ will make exactly one of $t+3$ , $t-3$ positive and the other negative, giving exactly two real $x$ -solutions.

Check the answer choices against the interval

Now compare each answer choice to the condition $-3 < t < 3$ :

$-6$ is less than $-3$ .
$-2$ is between $-3$ and $3$ .
$3$ is not strictly between $-3$ and $3$ ; it is an endpoint.
$8$ is greater than $3$ .

Only $t = -2$ lies in the interval $(-3, 3)$ , so it is the only choice that makes the equation have exactly two distinct real solutions.

Correct answer: $-2$ (choice B).

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Step-by-step Explanation

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Desmos Guide

Step-by-step Explanation