The equation is given. Let be the solution greater than of this equation. The value of can be written in the form , where is a positive integer. What is the value of ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 69 | Free SAT Prep | Aniko

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Question 69·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

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The equation

(x - 2)^4 - 20(x - 2)^2 + 64 = 0

is given. Let $x$ be the solution greater than $5$ of this equation. The value of $x$ can be written in the form $2 + \sqrt{k}$ , where $k$ is a positive integer. What is the value of $k$ ?

For equations where a shifted expression like $(x - a)$ appears with even powers (e.g., a fourth power and a square), treat the repeated part as a new variable to reduce the problem to a simple quadratic. Solve that quadratic carefully (checking the discriminant and division in the quadratic formula), then substitute back to solve for $x$ and apply any conditions (such as $x > 5$ ). Finally, match the required form—here, $2 + \sqrt{k}$ —to extract the requested value efficiently.

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See the hidden quadratic

Look at the repeated expression $(x - 2)$ : what happens if you let $u = (x - 2)^2$ ? How does the equation simplify in terms of $u$ ?

Solve for the new variable first

After you substitute $u = (x - 2)^2$ , you will get a standard quadratic equation in $u$ . Solve it using factoring or the quadratic formula.

Return to $x$ and apply the condition

Each value of $u$ gives you an equation $(x - 2)^2 = u$ . Solve these for $x$ , then use the condition that $x$ must be greater than $5$ to decide which $x$ to keep.

Match the required form

Once you have the correct $x$ , write it as $2 + \sqrt{k}$ and solve for $k$ by isolating the square root and squaring if needed.

Desmos Guide

Graph the original equation

In Desmos, enter y = (x - 2)^4 - 20(x - 2)^2 + 64. This will graph the function whose zeros are the solutions to the equation.

Find the solution greater than 5

Tap on the points where the graph crosses the x-axis (the x-intercepts). Identify the x-intercept that is greater than 5; this is the required solution for $x$ .

Compute the corresponding k

In a new Desmos line, enter (x_intercept - 2)^2, replacing x_intercept with the numeric value you found for the solution greater than 5. The output of this expression is the value of $k$ in $x = 2 + \sqrt{k}$ .

Step-by-step Explanation

Recognize the structure and substitute

Notice that the equation

(x - 2)^4 - 20(x - 2)^2 + 64 = 0

is a quadratic in $(x - 2)^2$ .

Let $u = (x - 2)^2$ . Then $(x - 2)^4$ becomes $u^2$ , so the equation turns into

u^2 - 20u + 64 = 0.

Solve the quadratic for $u$

Solve

u^2 - 20u + 64 = 0.

Use the quadratic formula:

u = \frac{20 \pm \sqrt{20^2 - 4 \cdot 1 \cdot 64}}{2}.

Compute the discriminant:

20^2 - 4 \cdot 64 = 400 - 256 = 144,

so $\sqrt{144} = 12$ . Then

\begin{aligned} u &= \frac{20 \pm 12}{2} \\ &= \frac{32}{2} \text{ or } \frac{8}{2} \\ &= 16 \text{ or } 4. \end{aligned}

So $u = 16$ or $u = 4$ .

Translate back to $x$

Remember $u = (x - 2)^2$ , so we have two equations:

$(x - 2)^2 = 16$
$(x - 2)^2 = 4$

Solve each by taking square roots.

For $(x - 2)^2 = 16$ :

x - 2 = \pm 4,

so $x = 2 + 4 = 6$ or $x = 2 - 4 = -2$ .

For $(x - 2)^2 = 4$ :

x - 2 = \pm 2,

so $x = 2 + 2 = 4$ or $x = 2 - 2 = 0$ .

All four solutions to the equation are $x = 6, -2, 4, 0$ .

Use the condition and find $k$

We are told $x$ is the solution greater than $5$ , so the only acceptable solution is

x = 6.

We must write $x$ in the form $2 + \sqrt{k}$ :

6 = 2 + \sqrt{k} \implies 6 - 2 = \sqrt{k} \implies 4 = \sqrt{k}.

Square both sides:

4^2 = k \implies k = 16.

So the value of $k$ is $\boxed{16}$ .

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Step-by-step Explanation

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