Simplify and rewrite the expression $\sqrt{x^2 - 10x + 25}$ without any radical symbols into an equivalent form that is mathematically valid for all real numbers $x$, paying specific attention to the interval where $x

Simplify and rewrite the expression $\sqrt{x^2 - 10x + 25}$ without any radical symbols into an equivalent form that is mathematically valid for all real numbers $x$ , paying specific attention to the interval where $x < 5$ .

Final answer

$|x - 5|$

Steps

Factor the expression inside the radical

To simplify the expression $\sqrt{x^2 - 10x + 25}$ for all real values of $x$ , we proceed in two main steps:

Notice that the radicand is a perfect square trinomial of the form $a^2 - 2ab + b^2 = (a - b)^2$ : $x^2 - 10x + 25 = (x - 5)^2$

Substituting this back into the radical gives: $\sqrt{x^2 - 10x + 25} = \sqrt{(x - 5)^2}$

Apply the principal square root identity

By definition, the principal (non-negative) square root of any squared real quantity $u$ is given by: $\sqrt{u^2} = |u|$

Applying this rule with $u = x - 5$ : $\sqrt{(x - 5)^2} = |x - 5|$

Common Pitfall: Why $x - 5$ is not valid for all real numbers

A common mistake is writing $\sqrt{(x - 5)^2} = x - 5$ .

However, the radical symbol $\sqrt{\phantom{x}}$ always denotes the non-negative square root. Consider the interval where $x < 5$ :

If $x = 2$ , then $x - 5 = 2 - 5 = -3$ .
Evaluating the original radical: $\sqrt{2^2 - 10(2) + 25} = \sqrt{4 - 20 + 25} = \sqrt{9} = 3$ .
If we used $x - 5$ , we would get $-3$ , which contradicts the definition of the principal square root.
The absolute value ensures the result remains non-negative: $|2 - 5| = |-3| = 3$ .

Thus, to remain mathematically valid for all real numbers $x$ (including when $x < 5$ , where the expression evaluates to $5 - x$ ), the expression must include absolute value bars: $|x - 5|$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the absolute value property of the principal square root and assumes √(a²) = a for all real numbers, concluding the expression is simply x - 5.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
x^2 - 10x + 25 = (x-5)^2 $\implies$ $\sqrt{x^2 - 10x + 25}$ = $\sqrt{(x-5)^2}$
Step 2
= x - 5
Correct
= |x - 5|
The principal square root $\sqrt{u^2}$ is defined as $|u|$ , not simply $u$ . When $x < 5$ , $x - 5$ is negative, so an absolute value is required to keep the expression non-negative for all real numbers.

WHAT YOU DID WELL

Compliments from Your Teacher

You factored the quadratic trinomial $x^2 - 10x + 25$ correctly into $(x - 5)^2$ .

You properly substituted the factored form back under the square root.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omission of absolute value on principal square root

Dropping the radical and exponent simultaneously as if $\sqrt{u^2} = u$ without considering negative values of $u$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating the square root and the square as straightforward inverse operations that cancel out directly, rather than applying the definition of the principal square root $\sqrt{u^2} = |u|$ which ensures non-negative outputs.

REASONING PATH

Your ACEpath teacher’s suggested approach

Principal Square Root Definition: For any real quantity $u$ , $\sqrt{u^2} = |u|$ .
Always write absolute value bars when simplifying $\sqrt{(\text{expression})^2}$ unless you already know the expression cannot be negative.
Test with a Boundary Value: The principal radical symbol $\sqrt{\phantom{x}}$ never produces a negative real number.
Pick a test value in the specified interval (e.g., $x = 0$ , where $x < 5$ ): $\sqrt{0^2 - 10(0) + 25} = \sqrt{25} = 5$ , whereas $0 - 5 = -5$ . Since $5 \neq -5$ , you need $|x - 5|$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the principal square root identity $\sqrt{u^2} = |u|$ to the radicand $(x - 5)^2$ .
Conclude that $\sqrt{(x - 5)^2} = |x - 5|$ , which evaluates to $-(x - 5) = 5 - x$ when $x < 5$ .

Final answer

$|x - 5|$

NEXT STEPS

Final Takeaways from Your Teacher

Review the identity $\sqrt{a^2} = |a|$ for real numbers.
Practice checking signs by substituting a test point when simplifying radicals with variables.
Solve expressions involving piecewise definitions or absolute values for intervals where the base is negative.

OverviewACEpath Teacher's Diagnosis

02 The student introduces the absolute value |x - 5| but incorrectly removes it for x < 5 by dropping the bars without negating the negative quantity.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\sqrt{x^2 - 10x + 25} = \sqrt{(x - 5)^2} = |x - 5|$
Step 2
$\text{When } x < 5,\ |x - 5| = x - 5$
Correct
$\text{When } x < 5,\ x - 5 < 0 \implies |x - 5| = -(x - 5) = 5 - x$
When $x < 5$ , the quantity inside the absolute value $(x - 5)$ is strictly negative. By definition, if $u < 0$ , then $|u| = -u$ . Dropping the bars directly yields a negative value, which contradicts the non-negative definition of absolute value.

WHAT YOU DID WELL

Compliments from Your Teacher

You factored the radicand into $(x - 5)^2$ completely and accurately.

You correctly identified that the principal square root of a squared real quantity requires absolute value bars, writing $|x - 5|$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Absolute Value Definition Error

Dropping absolute value bars without changing signs when the interior expression is negative.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You treated removing absolute value bars as simply deleting the vertical bars, forgetting that when an expression $u$ is negative ( $u < 0$ ), its absolute value is defined as the opposite of the expression, $|u| = -u$ .

REASONING PATH

Your ACEpath teacher’s suggested approach

Evaluate the Sign of the Interior: Before removing absolute value bars, test the sign of the inside expression: if $u \ge 0$ , then $|u| = u$ ; if $u < 0$ , then $|u| = -u$ .
Since $x < 5$ , subtract $5$ from both sides to see that $x - 5 < 0$ . Because the quantity is negative, replace $|x - 5|$ with $-(x - 5) = 5 - x$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

The expression valid for all real numbers $x$ is $|x - 5|$ .
To evaluate specifically on the interval $x < 5$ , note that $x < 5 \implies x - 5 < 0$ .
Apply the piecewise definition of absolute value for a negative argument: $|x - 5| = -(x - 5) = 5 - x$ .

Final answer

$|x - 5|$ (which equals $5 - x$ when $x < 5$ )

NEXT STEPS

Final Takeaways from Your Teacher

Practice substituting a specific test value (such as $x = 2$ ) whenever removing absolute value bars to verify that your output is strictly non-negative.
Review the formal piecewise definition of absolute value, specifically the case where the interior argument is less than zero.

OverviewACEpath Teacher's Diagnosis

03 The student distributes the square root across the individual terms of the radicand rather than factoring into a perfect square.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\sqrt{x^2 - 10x + 25}$ = $\sqrt{x^2}$ - $\sqrt{10x}$ + $\sqrt{25}$
Correct
$\sqrt{x^2 - 10x + 25}$ = $\sqrt{(x - 5)^2}$
Square roots cannot be distributed over addition or subtraction (that is, $\sqrt{A + B} \neq \sqrt{A} + \sqrt{B}$ ). Instead, recognize that $x^2 - 10x + 25$ is a perfect square trinomial equal to $(x - 5)^2$ .
Step 2
= x - $\sqrt{10x}$ + 5
Correct
= |x - 5|
This result follows directly from the invalid distribution in Step 1. Taking the principal square root of the squared quantity $(x - 5)^2$ gives $|x - 5|$ , which evaluates to $5 - x$ when $x < 5$ .

WHAT YOU DID WELL

Compliments from Your Teacher

You recognized that terms like $x^2$ and $25$ have familiar individual square roots.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Distributing radical over addition/subtraction

You split the radical term by term across addition and subtraction signs instead of treating the entire radicand as a single composite quantity.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating radicals as linear operators that distribute over sums and differences (confusing the multiplication property $\sqrt{ab} = \sqrt{a}\sqrt{b}$ with addition $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

Radical Addition Rule: $\sqrt{A \pm B} \neq \sqrt{A} \pm \sqrt{B}$ . Radicals distribute only across products and quotients of non-negative numbers.
Never split a radical across plus or minus signs. Look for a common factor or factor into a perfect square trinomial first.
Principal Square Root Identity: For any real expression $u$ , $\sqrt{u^2} = |u|$ .
Always write absolute value bars around the base when simplifying $\sqrt{(expression)^2}$ unless the base is guaranteed to be non-negative.

CORRECTED PATH

ACEpath Teacher's Demonstration

Factor the perfect square trinomial inside the radical: $x^2 - 10x + 25 = (x - 5)^2$ .
Rewrite the whole expression: $\sqrt{(x - 5)^2}$ .
Apply the identity $\sqrt{u^2} = |u|$ with $u = x - 5$ to ensure non-negativity for all real $x$ .

Final answer

|x - 5|

NEXT STEPS

Final Takeaways from Your Teacher

Review the distinction between radical multiplication rules $\sqrt{ab} = \sqrt{a}\sqrt{b}$ and the non-distributive nature of addition $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$ .
Practice factoring quadratic expressions of the form $a^2 - 2ab + b^2 = (a - b)^2$ before applying radical simplification.
Remember to apply absolute value bars when evaluating $\sqrt{u^2} = |u|$ for unrestricted real variables.

Lock it in

Question

For any real number $a$ , what is the exact identity for $\sqrt{a^2}$ ?

Answer

$\sqrt{a^2} = |a|$ . By definition, the radical symbol $\sqrt{\phantom{x}}$ denotes the principal (non-negative) square root.

Resources

Answer and steps

Final answer

Steps

Common Pitfall: Why x−5x - 5x−5 is not valid for all real numbers

Typical errors

01 The student ignores the absolute value property of the principal square root and assumes √(a²) = a for all real numbers, concluding the expression is simply x - 5.

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 introduces the absolute value |x - 5| but incorrectly removes it for x < 5 by dropping the bars without negating the negative quantity.

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 distributes the square root across the individual terms of the radicand rather than factoring into a perfect square.

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

Common Pitfall: Why $x - 5$ is not valid for all real numbers