A student attempts to solve the equation $\frac{x^2 - 16}{x - 4} = 8$ by factoring the numerator as $(x - 4)(x + 4)$, canceling the factor $(x - 4)$ to obtain $x + 4 = 8$, and concluding that $x = 4$ is the unique solution. Determine the true number of solutions to the original equation and identify why the student's conclusion fails.

A student attempts to solve the equation $\frac{x^2 - 16}{x - 4} = 8$ by factoring the numerator as $(x - 4)(x + 4)$ , canceling the factor $(x - 4)$ to obtain $x + 4 = 8$ , and concluding that $x = 4$ is the unique solution. Determine the true number of solutions to the original equation and identify why the student's conclusion fails.

Final answer

No solution. (The expression is undefined at $x = 4$ due to division by zero; the domain is $\{x \in \mathbb{R} \mid x \neq 4\}$ . Since the only candidate root is $x = 4$ , the solution set is empty $\emptyset$ .)

Steps

Identify the domain restriction of the equation

Before simplifying any rational expression, check for values of the variable that make any denominator equal to zero, because division by zero is mathematically undefined.

In the equation: $\frac{x^2 - 16}{x - 4} = 8$

The denominator is $x - 4$ . Setting it to zero gives: $x - 4 = 0 \implies x = 4$

Therefore, the equation is only defined for $x \neq 4$ . The domain is all real numbers except $4$ : $\text{Domain: } \{x \in \mathbb{R} \mid x \neq 4\}$

Solve the simplified equation

Under the condition that $x \neq 4$ , the term $(x - 4)$ is non-zero, which makes dividing both the numerator and the denominator by $(x - 4)$ valid: $\frac{(x - 4)(x + 4)}{x - 4} = 8$ $x + 4 = 8$

Subtract $4$ from both sides: $x = 4$

This yields $x = 4$ as the only candidate root.

Compare the candidate solution with the domain

Now verify whether the candidate root satisfies the original equation:

The candidate solution is $x = 4$ .
However, substituting $x = 4$ into the original expression gives: $\frac{4^2 - 16}{4 - 4} = \frac{0}{0}$ which is undefined.

Thus, $x = 4$ is an extraneous solution introduced by canceling a factor that equals zero.

Conclusion

True number of solutions: No solution (the solution set is empty, $\emptyset$ ).
Why the student's conclusion fails: The student canceled the common factor $(x - 4)$ without accounting for the domain restriction $x \neq 4$ . Canceling a factor of zero is invalid, making the resulting value an extraneous root rather than a valid solution.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student concludes there is one solution by accepting x = 4 after only checking that it satisfies the simplified equation x + 4 = 8 rather than the original equation.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
There is 1 solution.
Correct
There are no solutions ( $0$ solutions).
The original equation is undefined at $x = 4$ due to division by zero. Since $x = 4$ is the only candidate root, the solution set is empty.
Step 2
The student's conclusion does not fail because substituting 4 gives 4 + 4 = 8, which is true.
Correct
The student's conclusion fails because canceling the factor $(x - 4)$ is only valid when $x - 4 \neq 0$ ( $x \neq 4$ ). Substituting $x = 4$ into the original equation produces the undefined expression $\frac{0}{0}$ .
Checking the simplified equation $x + 4 = 8$ is not sufficient because canceling $(x - 4)$ assumes $x \neq 4$ . Substituting $x = 4$ into the original denominator results in division by zero, making $x = 4$ an extraneous root.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly verified that $x = 4$ satisfies the simplified equation $x + 4 = 8$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Extraneous Solution / Ignored Domain Restriction

Testing the candidate root in the simplified equation rather than checking the domain of the original rational expression.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated algebraic cancellation as an unconditional equivalence, overlooking that dividing numerator and denominator by $(x - 4)$ is valid only when $x \neq 4$ . Consequently, substituting the candidate into the simplified form fails to detect that it causes division by zero in the original expression.

REASONING PATH

Your ACEpath teacher’s suggested approach

Establish the Domain First: For any rational expression $\frac{P(x)}{Q(x)}$ , division by zero is undefined, so $Q(x) \neq 0$ .
Set every denominator equal to $0$ at the start and list all excluded values before factoring or canceling.
Always Test the Original Equation: Candidate roots must satisfy the original equation, not just an intermediate or simplified version.
Substitute any candidate root back into the original unsimplified denominator to confirm it does not produce division by zero.

CORRECTED PATH

ACEpath Teacher's Demonstration

Set the denominator to zero: $x - 4 = 0 \implies x = 4$ . Therefore, the domain restriction is $x \neq 4$ .
For $x \neq 4$ , cancel the common non-zero factor: $\frac{(x - 4)(x + 4)}{x - 4} = 8 \implies x + 4 = 8 \implies x = 4$ .
Compare the candidate root $x = 4$ with the domain restriction $x \neq 4$ . Since $x = 4$ is excluded, it is an extraneous solution.

Final answer

The equation has no solution ( $0$ solutions). The student's conclusion fails because substituting $x = 4$ into the original expression causes division by zero ( $\frac{0}{0}$ ), making $x = 4$ an extraneous root.

NEXT STEPS

Final Takeaways from Your Teacher

State domain restrictions explicitly as the very first step whenever an equation contains variable denominators.
Cross-check every candidate root against your excluded-values list before writing down the final solution count.
Practice identifying removable discontinuities (holes) versus valid solutions in rational equations.

OverviewACEpath Teacher's Diagnosis

02 The student confuses an undefined expression with having infinitely many solutions by interpreting 0/0 as an identity.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
At $x = 4$ the equation gives $\frac{0}{0}$
Step 2
which means any value works rather than having a unique solution
Correct
Division by zero is undefined, so $x = 4$ is excluded from the domain ( $\{x \in \mathbb{R} \mid x \neq 4\}$ )
The form $\frac{0}{0}$ does not mean every real number satisfies the equation. Division by zero is undefined in real arithmetic, which makes $x = 4$ an extraneous candidate rather than an identity.
Step 3
There are infinitely many solutions.
Correct
There is no solution (the solution set is empty, $\emptyset$ )
Since the only candidate root $x = 4$ is excluded by the domain restriction $x \neq 4$ , no value satisfies the original equation.

WHAT YOU DID WELL

Compliments from Your Teacher

You checked the candidate root $x = 4$ in the original equation.

You recognized that substituting $x = 4$ yields $\frac{0}{0}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Undefined Expression vs Identity Misconception

Interpreting an undefined division-by-zero expression $\frac{0}{0}$ as an identity that allows infinitely many solutions.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You confused the undefined result of dividing by zero (which excludes the input from the equation's domain) with an algebraic identity like $0x = 0$ (which is satisfied by all real numbers).

REASONING PATH

Your ACEpath teacher’s suggested approach

Denominator Zero Rule: Any value of $x$ that makes a denominator zero cannot be part of the domain, because division by zero is undefined.
Always establish domain restrictions before simplifying by setting each denominator $\neq 0$ .
Extraneous Solution Check: If simplifying an equation produces a root that violates the original domain restriction, that root is extraneous and must be discarded.
Compare every candidate solution against the domain before finalizing the count of solutions.

CORRECTED PATH

ACEpath Teacher's Demonstration

State the domain of the equation $\frac{x^2 - 16}{x - 4} = 8$ by requiring $x - 4 \neq 0$ , which gives $x \neq 4$ .
For $x \neq 4$ , simplify $\frac{(x - 4)(x + 4)}{x - 4} = 8$ to obtain $x + 4 = 8$ , which yields the candidate $x = 4$ .
Compare the candidate root $x = 4$ with the domain restriction $x \neq 4$ : since $4$ is excluded, $x = 4$ is an extraneous root.

Final answer

No solution

NEXT STEPS

Final Takeaways from Your Teacher

State domain restrictions upfront by setting every denominator non-zero before factoring or canceling.
Remember that $\frac{0}{0}$ signifies an undefined value in an algebraic equation, never an identity that allows all inputs.
Discard any candidate solution that falls outside the domain to determine the true number of solutions.

Lock it in

Question

When can you legally cancel a factor $(x - a)$ from both the numerator and denominator of an algebraic fraction?

Answer

Only when $x \neq a$ . Canceling $(x - a)$ is dividing by that factor, which is invalid if the factor equals zero.

Resources

Answer and steps

Final answer

Steps

Conclusion

Typical errors

01 The student concludes there is one solution by accepting x = 4 after only checking that it satisfies the simplified equation x + 4 = 8 rather than the original equation.

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 confuses an undefined expression with having infinitely many solutions by interpreting 0/0 as an identity.

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

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