Consider the expression $P(x) = \left( x^{1/2} + x^{-1/2} \right)^2 - \left( x^{1/2} - x^{-1/2} \right)^2$ for $x > 0$. Although a student expects the result to depend heavily on the variable $x$ as $x \to 0^+$ or $x \to \infty$, simplify the expression algebraically and state its exact value.

Consider the expression $P(x) = \left( x^{1/2} + x^{-1/2} \right)^2 - \left( x^{1/2} - x^{-1/2} \right)^2$ for $x > 0$ . Although a student expects the result to depend heavily on the variable $x$ as $x \to 0^+$ or $x \to \infty$ , simplify the expression algebraically and state its exact value.

Final answer

$4$

Steps

Recall exponent rules and algebraic identities

To simplify the expression $P(x)$ , we can expand both squared binomials using standard algebraic identities.

For any positive base $x$ and real exponents, the product rule for exponents states: $x^a \cdot x^b = x^{a+b}$

Applying this to our terms: $x^{1/2} \cdot x^{-1/2} = x^{1/2 + (-1/2)} = x^0 = 1$

Also, applying the power of a power rule $(x^a)^b = x^{ab}$ : $(x^{1/2})^2 = x^{(1/2) \cdot 2} = x^1 = x$ $(x^{-1/2})^2 = x^{(-1/2) \cdot 2} = x^{-1} = \frac{1}{x}$

Expand each squared term

Recall the expansion formulas $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ .

Let $a = x^{1/2}$ and $b = x^{-1/2}$ :

Expand the first term: $\left( x^{1/2} + x^{-1/2} \right)^2 = (x^{1/2})^2 + 2(x^{1/2})(x^{-1/2}) + (x^{-1/2})^2$ $\left( x^{1/2} + x^{-1/2} \right)^2 = x + 2(1) + x^{-1} = x + 2 + x^{-1}$
Expand the second term: $\left( x^{1/2} - x^{-1/2} \right)^2 = (x^{1/2})^2 - 2(x^{1/2})(x^{-1/2}) + (x^{-1/2})^2$ $\left( x^{1/2} - x^{-1/2} \right)^2 = x - 2(1) + x^{-1} = x - 2 + x^{-1}$

Subtract the two expressions

Substitute the expanded forms back into $P(x)$ : $P(x) = (x + 2 + x^{-1}) - (x - 2 + x^{-1})$

Distribute the negative sign carefully: $P(x) = x + 2 + x^{-1} - x + 2 - x^{-1}$

Combine like terms: $P(x) = (x - x) + (x^{-1} - x^{-1}) + (2 + 2)$ $P(x) = 0 + 0 + 4 = 4$

Alternative Shortcut

We can directly recognize the identity for the difference of two squares of a sum and difference: $(a + b)^2 - (a - b)^2 = 4ab$ Setting $a = x^{1/2}$ and $b = x^{-1/2}$ : $P(x) = 4(x^{1/2})(x^{-1/2}) = 4(x^0) = 4(1) = 4$

Common Pitfall

A frequent error is incorrectly evaluating powers such as $(x^{1/2})^2$ as $x^{1/4}$ instead of multiplying exponents to get $x^1$ , or miscalculating the cross-term product $x^{1/2} \cdot x^{-1/2}$ rather than recognizing that reciprocal powers sum to an exponent of zero.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets the cross-terms when expanding the squares, treating (a + b)² as a² + b².

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
P(x) = (x^{1/2} + x^{-1/2})^2 - (x^{1/2} - x^{-1/2})^2 = (x + x^{-1}) - (x + x^{-1})
Correct
(x + 2 + x^{-1}) - (x - 2 + x^{-1})
When expanding a binomial square $(a \pm b)^2$ , you must include the middle term $\pm 2ab$ . Here, $2(x^{1/2})(x^{-1/2}) = 2(x^0) = 2$ , so the expansions should be $x + 2 + x^{-1}$ and $x - 2 + x^{-1}$ .
Step 2
= 0
Correct
= 4
Subtracting your expanded expressions correctly yielded $0$ , but this result is affected by the missing cross-terms from Step 1.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly transcribed the original expression.

You correctly applied the power of a power rule to obtain $(x^{1/2})^2 = x$ and $(x^{-1/2})^2 = x^{-1}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted cross-terms in binomial expansion

You distributed the exponent $2$ directly to each term inside the parentheses rather than multiplying the binomial by itself.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating $(a \pm b)^2$ as $a^2 \pm b^2$ (or $a^2 + b^2$ ) fails to account for the cross products generated when expanding a polynomial.

REASONING PATH

Your ACEpath teacher’s suggested approach

Binomial Square Identity: $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$
Always write out all three terms: square of the first, twice the product of both, and square of the second.
Difference of Squares of Sum and Difference: $(a + b)^2 - (a - b)^2 = 4ab$
Notice that the squared terms $a^2$ and $b^2$ cancel completely upon subtraction, leaving only the difference between $+2ab$ and $-2ab$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the identity $(a \pm b)^2 = a^2 \pm 2ab + b^2$ with $a = x^{1/2}$ and $b = x^{-1/2}$ :
$\left(x^{1/2} + x^{-1/2}\right)^2 = x + 2(x^{1/2})(x^{-1/2}) + x^{-1} = x + 2 + x^{-1}$
$\left(x^{1/2} - x^{-1/2}\right)^2 = x - 2(x^{1/2})(x^{-1/2}) + x^{-1} = x - 2 + x^{-1}$
Subtract the second expansion from the first:
$P(x) = (x + 2 + x^{-1}) - (x - 2 + x^{-1}) = x - x + 2 - (-2) + x^{-1} - x^{-1} = 4$

Final answer

NEXT STEPS

Final Takeaways from Your Teacher

Review the expansion of binomials $(a \pm b)^2$ to build the habit of writing all three terms including $\pm 2ab$ .
Practice using the shortcut $(a + b)^2 - (a - b)^2 = 4ab$ to simplify similar symmetric expressions directly.

OverviewACEpath Teacher's Diagnosis

02 The student fails to distribute the negative sign to the middle term when subtracting the second expanded expression.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
P(x) = (x + 2 + x^{-1}) - (x - 2 + x^{-1})
Step 2
= x + 2 + x^{-1} - x - 2 - x^{-1}
Correct
= x + 2 + x^{-1} - x + 2 - x^{-1}
Distributing the minus sign across $-(x - 2 + x^{-1})$ changes the sign of each term inside the parentheses: $-(-2)$ must become $+2$ , not $-2$ .
Step 3
= 0
Correct
= 4
This final value follows directly from the negative sign error in the previous step.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly expanded $(x^{1/2} + x^{-1/2})^2$ to $x + 2 + x^{-1}$ by recognizing that $2(x^{1/2})(x^{-1/2}) = 2(x^0) = 2$ .

You correctly expanded $(x^{1/2} - x^{-1/2})^2$ to $x - 2 + x^{-1}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Sign distribution error across parentheses

You distributed the leading negative sign to $x$ and $x^{-1}$ , but missed flipping the sign of $-2$ to $+2$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Incomplete distribution of a negative sign over a polynomial when removing grouping symbols containing negative terms.

REASONING PATH

Your ACEpath teacher’s suggested approach

Distribute negative signs to every term: $-(A - B + C) = -A + B - C$
Always multiply every term inside the parentheses by $-1$ , paying close attention to terms that are already negative.

CORRECTED PATH

ACEpath Teacher's Demonstration

P(x) = x + 2 + x^{-1} - x + 2 - x^{-1}
P(x) = (x - x) + (x^{-1} - x^{-1}) + (2 + 2)

Final answer

NEXT STEPS

Final Takeaways from Your Teacher

Review sign distribution rules when clearing parentheses preceded by a minus sign.
Practice using the identity $(a + b)^2 - (a - b)^2 = 4ab$ as a quick alternative check.

OverviewACEpath Teacher's Diagnosis

03 The student incorrectly evaluates the product of reciprocal powers x^(1/2) · x^(-1/2) as x instead of 1.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
(x^{ $\frac{1}{2}$ } + x^{- $\frac{1}{2}$ })^2 = x + 2x + x^{-1} $\implies$ (x^{ $\frac{1}{2}$ } - x^{- $\frac{1}{2}$ })^2 = x - 2x + x^{-1}
Correct
(x^{1/2} + x^{-1/2})^2 = x + 2 + x^{-1} $\implies$ (x^{1/2} - x^{-1/2})^2 = x - 2 + x^{-1}
When expanding $(a \pm b)^2$ , the cross term involves $2ab = 2(x^{1/2})(x^{-1/2})$ . By the product rule for exponents, $x^{1/2} \cdot x^{-1/2} = x^{1/2 + (-1/2)} = x^0 = 1$ , so the cross terms are $+2$ and $-2$ , not $+2x$ and $-2x$ .
Step 2
P(x) = (3x + x^{-1}) - (-x + x^{-1})
Correct
P(x) = (x + 2 + x^{-1}) - (x - 2 + x^{-1})
You correctly collected your previous terms and set up the subtraction, but the grouped expressions inherited the incorrect cross terms from Step 1.
Step 3
= 4x
Correct
= 4
Carrying through your algebraic subtraction gave $3x - (-x) = 4x$ , but with the correct constant middle terms, $(2) - (-2) = 4$ while the $x$ and $x^{-1}$ terms completely cancel.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly recognized and applied the algebraic expansion form $(a \pm b)^2 = a^2 \pm 2ab + b^2$ .

You accurately computed the powers of powers $(x^{1/2})^2 = x$ and $(x^{-1/2})^2 = x^{-1}$ .

You distributed the negative sign across the second grouped binomial and combined like terms accurately according to your earlier lines.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Product of Reciprocal Powers Miscalculated

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated the product of reciprocal bases $x^{1/2} \cdot x^{-1/2}$ as $x$ rather than adding the exponents $1/2 + (-1/2) = 0$ to get $x^0 = 1$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Reciprocal Exponent Product Rule: $x^a \cdot x^{-a} = x^{a - a} = x^0 = 1$ for any $x > 0$ .
Whenever you multiply terms with opposite exponents, add the exponents to get zero, which simplifies to the constant $1$ .
Difference of Squares Shortcut: $(a + b)^2 - (a - b)^2 = 4ab$ .
Use the identity directly to avoid expanding all six terms: $4(x^{1/2})(x^{-1/2}) = 4(1) = 4$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Expand the first binomial: $(x^{1/2} + x^{-1/2})^2 = x + 2(x^{1/2} \cdot x^{-1/2}) + x^{-1} = x + 2 + x^{-1}$ .
Expand the second binomial: $(x^{1/2} - x^{-1/2})^2 = x - 2(x^{1/2} \cdot x^{-1/2}) + x^{-1} = x - 2 + x^{-1}$ .
Subtract the two expressions: $P(x) = (x + 2 + x^{-1}) - (x - 2 + x^{-1}) = (x - x) + (2 - (-2)) + (x^{-1} - x^{-1}) = 4$ .

Final answer

NEXT STEPS

Final Takeaways from Your Teacher

Practice multiplying reciprocal powers such as $a^k \cdot a^{-k}$ to solidify that the product equals $1$ , not $a$ .
Review the special identity $(a + b)^2 - (a - b)^2 = 4ab$ as an efficient shortcut for expressions of this structure.

Lock it in

Question

What are the full expansions of $(a + b)^2$ and $(a - b)^2$ ?

Answer

$(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ . Never forget the middle cross-term $\pm 2ab$ .

Resources

Answer and steps

Final answer

Steps

Alternative Shortcut

Common Pitfall

Typical errors

01 The student forgets the cross-terms when expanding the squares, treating (a + b)² as a² + b².

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 fails to distribute the negative sign to the middle term when subtracting the second expanded expression.

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 incorrectly evaluates the product of reciprocal powers x^(1/2) · x^(-1/2) as x instead of 1.

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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