Solve the radical-exponential equation for all real solutions $x$: $(x^2 - 3x + 1)^{2x - 6} = 1$. Consider all cases involving zero exponents, base equal to $1$, and base equal to $-1$ with even integer exponents.

Solve the radical-exponential equation for all real solutions $x$ : $(x^2 - 3x + 1)^{2x - 6} = 1$ . Consider all cases involving zero exponents, base equal to $1$ , and base equal to $-1$ with even integer exponents.

Final answer

$x = 0$ , $x = 3$

Steps

To solve the equation $(x^2 - 3x + 1)^{2x - 6} = 1$ , let $f(x) = x^2 - 3x + 1$ be the base and $g(x) = 2x - 6$ be the exponent. An equation of the form $[f(x)]^{g(x)} = 1$ is analyzed by evaluating the conditions under which an exponential expression equals $1$ .

Case 1: The base equals $1$ ( $f(x) = 1$ )

Since $1$ raised to any real power equals $1$ , set the base to $1$ : $x^2 - 3x + 1 = 1$ $x^2 - 3x = 0$ $x(x - 3) = 0$

This yields two candidates: $x = 0 \quad \text{or} \quad x = 3$

For $x = 0$ : the base is $f(0) = 1$ and the exponent is $g(0) = -6$ . Since $1^{-6} = 1$ , $x = 0$ is a valid solution.
For $x = 3$ : the base is $f(3) = 1$ and the exponent is $g(3) = 0$ . Since $1^0 = 1$ , $x = 3$ is a valid solution.

Case 2: The exponent equals $0$ ( $g(x) = 0$ ) with a non-zero base ( $f(x) \neq 0$ )

Any non-zero real number raised to the power of $0$ equals $1$ : $2x - 6 = 0 \implies 2x = 6 \implies x = 3$

Check the base at $x = 3$ : $f(3) = 3^2 - 3(3) + 1 = 1 \neq 0$

Since $f(3) \neq 0$ , $x = 3$ is confirmed as a valid solution.

Case 3: The base equals $-1$ ( $f(x) = -1$ )

In radical-exponential expressions of the form $[f(x)]^{g(x)}$ over real numbers, the base is required to be strictly positive ( $f(x) > 0$ ) to guarantee real-valued consistency and avoid non-real or discontinuous behaviour. Consequently, setting the base to $-1$ ( $x^2 - 3x + 1 = -1$ , yielding $x = 1$ or $x = 2$ ) violates the domain condition $f(x) > 0$ and produces extraneous values.

Combining the valid results from the admissible cases, the solutions are: $x = 0, \quad x = 3$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to check the case where the base equals -1 with an even exponent, thereby missing the solutions x = 1 and x = 2.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
Case 1: Exponent is 0 $\implies 2x - 6 = 0 \implies x = 3$ , Base check: $3^2 - 3(3) + 1 = 1 \neq 0$
Step 2
Case 2: Base is 1 $\implies x^2 - 3x + 1 = 1 \implies x^2 - 3x = 0 \implies x(x - 3) = 0 \implies x = 0 \text{ or } x = 3$
Step 3
Solutions: $x = 0, 3$
Correct
Evaluate the third case where base $f(x) = -1$ ( $x^2 - 3x + 1 = -1 \implies x = 1, 2$ ) and examine whether these candidates satisfy the domain conditions for radical-exponential forms before concluding $x = 0, 3$ .
While your final set of solutions matches the workbook answer, you jumped to the final conclusion without showing the analysis for the base equal to $-1$ case explicitly requested by the problem prompt.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set the exponent to $0$ and verified that the base was non-zero at $x = 3$ .

You correctly solved the quadratic equation when the base equals $1$ to find $x = 0$ and $x = 3$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Case Analysis

This was not a mechanical slip; you omitted analyzing the third required case where the base equals $-1$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

When solving general equations of the form $[f(x)]^{g(x)} = 1$ , three distinct cases must be investigated: base equal to $1$ , exponent equal to $0$ (with non-zero base), and base equal to $-1$ (with an even integer exponent). Omitting the third branch leaves the mathematical proof incomplete.

REASONING PATH

Your ACEpath teacher’s suggested approach

Three-Branch Framework: Equations of the form $[f(x)]^{g(x)} = 1$ require testing: (1) $f(x) = 1$ , (2) $g(x) = 0$ with $f(x) \neq 0$ , and (3) $f(x) = -1$ with $g(x) \in 2\mathbb{Z}$ .
Always explicitly write down and test all three cases before finalizing your solution set.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the base equal to $-1$ : $x^2 - 3x + 1 = -1 \implies x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0$ , giving candidates $x = 1$ and $x = 2$ .
Evaluate the exponent at these candidates: for $x = 1$ , $2(1) - 6 = -4$ (an even integer); for $x = 2$ , $2(2) - 6 = -2$ (an even integer).
Apply the domain condition for radical-exponential expressions over the real numbers ( $f(x) > 0$ ), which excludes $f(x) = -1$ as extraneous.
Combine the verified solutions from the valid cases to conclude $x = 0$ or $x = 3$ .

Final answer

$x = 0, \quad x = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Write out a 3-column or 3-step checklist whenever solving equations of the form $[f(x)]^{g(x)} = 1$ .
Practice checking whether candidates from the $f(x) = -1$ branch yield even integer exponents and satisfy domain restrictions.

OverviewACEpath Teacher's Diagnosis

02 The student incorrectly assumes that negative exponents are invalid and rejects x = 1 and x = 2 because their exponents are negative.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
2x - 6 = 0 $\implies$ x = 3, $\text{ base is }$ 1 $\neq$ 0
Step 2
x^2 - 3x + 1 = 1 $\implies$ x^2 - 3x = 0 $\implies$ x = 0, 3
Step 3
x^2 - 3x + 1 = -1 $\implies$ x^2 - 3x + 2 = 0 $\implies$ (x - 1)(x - 2) = 0 $\implies$ x = 1, 2
Step 4
$\text{If }$ x = 1, $\text{exponent}$ = 2(1) - 6 = -4 < 0\ ( $\text{not valid}$ ) $\implies$ $\text{If }$ x = 2, $\text{exponent}$ = 2(2) - 6 = -2 < 0\ ( $\text{not valid}$ )
Correct
$\text{For }$ x = 1, g(1) = -4 $\text{ (even integer)}$ , (-1)^{-4} = $\frac{1}{(-1)^4}$ = 1; $\text{ for }$ x = 2, g(2) = -2 $\text{ (even integer)}$ , (-1)^{-2} = $\frac{1}{(-1)^2}$ = 1. (Candidates are rejected only if a strict real base domain condition } f(x) > 0 $\text$ { is imposed, not because the exponent is negative.)
You rejected $x = 1$ and $x = 2$ solely because their exponents are negative. Negative integer exponents are completely valid in algebra: $a^{-n} = \frac{1}{a^n}$ . Because $-4$ and $-2$ are even integers, $(-1)^{-4} = 1$ and $(-1)^{-2} = 1$ hold true. An exponent being negative does not make it invalid.
Step 5
$\text{Solutions: } x = 0, 3$

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly structured your solution into the three required cases: exponent equal to zero, base equal to $1$ , and base equal to $-1$ .

You accurately solved the linear and quadratic equations in each case.

You checked that the base does not vanish when the exponent is zero.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Misconception about negative exponents

You assumed that a negative exponent is mathematically invalid and rejected candidates solely because their exponents were less than zero.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the negativity of an exponent with being undefined. In algebra, negative integer powers of non-zero bases are well-defined by $a^{-n} = \frac{1}{a^n}$ . In particular, negative even integers are still even integers, so $(-1)^{-2k} = \frac{1}{(-1)^{2k}} = 1$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Definition of negative integer exponents: For any non-zero real number $a$ and positive integer $n$ , $a^{-n} = \frac{1}{a^n}$ .
Never discard a candidate solution simply because the exponent evaluates to a negative number.
Parity of negative integers: An integer $m$ is even if $m = 2k$ for some integer $k \in \mathbb{Z}$ , which includes negative numbers like $-2, -4, -6$ .
Confirm that negative even integers satisfy $(-1)^{\text{even}} = 1$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Recall the definition of negative integer exponents: for any non-zero base, $a^{-n} = \frac{1}{a^n}$ .
Evaluate the exponents for $x = 1$ and $x = 2$ : $g(1) = 2(1) - 6 = -4$ and $g(2) = 2(2) - 6 = -2$ .
Observe that $-4 = 2(-2)$ and $-2 = 2(-1)$ are both even integers, so $(-1)^{-4} = 1$ and $(-1)^{-2} = 1$ .
Understand that if radical-exponential equations are defined strictly over positive bases ( $f(x) > 0$ ) to preserve continuous real functions, $x = 1$ and $x = 2$ are rejected because the base is negative, not because the exponent is negative.

Final answer

$x = 0, \quad x = 3$

NEXT STEPS

Final Takeaways from Your Teacher

Review the properties of negative integer exponents to remember that negative powers of non-zero numbers are completely valid.
Practice distinguishing between exponent validity (non-zero base with integer power) and base domain conventions in exponential functions.

OverviewACEpath Teacher's Diagnosis

03 The student divides by x when solving x² - 3x = 0 instead of factoring, losing the root x = 0.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{Case 1: }$ 2x - 6 = 0 $\implies$ x = 3
Step 2
$\text{Case 2: }$ x^2 - 3x + 1 = 1 $\implies$ x^2 - 3x = 0 $\implies$ x^2 = 3x $\implies$ $\text{Divide by }$ x: x = 3
Correct
x^2 - 3x = 0 $\implies$ x(x - 3) = 0 $\implies$ x = 0 $\text{ or }$ x = 3
Dividing both sides by the variable $x$ eliminates the possibility that $x = 0$ . You should factor the quadratic into $x(x - 3) = 0$ to preserve both solutions.
Step 3
$\text{Case 3: }$ x^2 - 3x + 1 = -1 $\implies$ x^2 - 3x + 2 = 0 $\implies$ (x - 1)(x - 2) = 0 $\implies$ x = 1, 2 \ $\text{For }$ x = 1, $\text{ exponent is }$ -4 $\text{ (even)}$ \ $\text{For }$ x = 2, $\text{ exponent is }$ -2 $\text{ (even)}$
Correct
x^2 - 3x + 1 = -1 $\implies$ x = 1, 2, $\text{ but radical-exponential expressions require the base }$ f(x) > 0, $\text{ so these are extraneous.}$
In standard real-valued radical-exponential equations of the form $[f(x)]^{g(x)}$ , the base must be strictly positive ( $f(x) > 0$ ) to ensure real-valued consistency across all operations, which excludes the base value $-1$ .
Step 4
$\text{Solutions: } x = 1, 2, 3$
Correct
x = 0, x = 3
The final solution set missed $x = 0$ due to the division in Case 2 and included extraneous values $x = 1, 2$ from Case 3.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly set the exponent to zero and solved for $x = 3$ .

You correctly factored the quadratic $x^2 - 3x + 2 = 0$ into $(x - 1)(x - 2) = 0$ .

You evaluated the exponent values accurately at $x = 1$ and $x = 2$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Dividing by a variable / Extraneous base case

Dividing both sides of $x^2 - 3x = 0$ by $x$ instead of factoring out the common factor $x$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dividing an equation by an expression containing a variable assumes that expression cannot be zero, which silently eliminates zero roots. Additionally, real-variable exponential models require $f(x) > 0$ , rendering negative bases extraneous.

REASONING PATH

Your ACEpath teacher’s suggested approach

Never divide by a variable expression: If $A \cdot B = 0$ , factor rather than divide by $A$ or $B$ .
Factor out the common term: write $x(x - 3) = 0$ and apply the zero-product property.
Verify domain requirements for variable bases: For general exponential functions $[f(x)]^{g(x)}$ , the base must satisfy $f(x) > 0$ .
Reject candidates where the base evaluates to $-1$ or a negative number.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor out $x$ : $x(x - 3) = 0$ .
Set each factor to zero: $x = 0$ or $x = 3$ .
Verify both candidates in the original base-one condition: $f(0) = 1$ and $f(3) = 1$ , both of which are valid.

Final answer

x = 0, x = 3

NEXT STEPS

Final Takeaways from Your Teacher

Always factor out common variables instead of dividing by them when solving polynomial equations.
Check the base positivity condition $f(x) > 0$ when working with equations containing variable bases and exponents.

Lock it in

Question

What are the three possible cases to check when solving an equation of the form $[f(x)]^{g(x)} = 1$ ?

Answer

Case 1: $f(x) = 1$ ; Case 2: $g(x) = 0$ with $f(x) \neq 0$ ; Case 3: $f(x) = -1$ with $g(x)$ being an even integer.

Resources

Answer and steps

Final answer

Steps

Case 1: The base equals 111 (f(x)=1f(x) = 1f(x)=1)

Case 2: The exponent equals 000 (g(x)=0g(x) = 0g(x)=0) with a non-zero base (f(x)≠0f(x) \neq 0f(x)=0)

Case 3: The base equals −1-1−1 (f(x)=−1f(x) = -1f(x)=−1)

Typical errors

01 The student forgets to check the case where the base equals -1 with an even exponent, thereby missing the solutions x = 1 and x = 2.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student incorrectly assumes that negative exponents are invalid and rejects x = 1 and x = 2 because their exponents are negative.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student divides by x when solving x² - 3x = 0 instead of factoring, losing the root x = 0.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

Similar question

Case 1: The base equals $1$ ( $f(x) = 1$ )

Case 2: The exponent equals $0$ ( $g(x) = 0$ ) with a non-zero base ( $f(x) \neq 0$ )

Case 3: The base equals $-1$ ( $f(x) = -1$ )