Solve the Square Root Expression: Simplifying √(36x⁴)

Square Root Simplification with Variables

Solve the following exercise:

36x4= \sqrt{36x^4}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Simplify the following expression
00:03 The square root of number (A) multiplied by the square root of another number (B)
00:07 Equals the square root of their product (A times B)
00:10 Apply this formula to our exercise, and convert from root 1 to two
00:15 Factor 36 into 6 squared
00:21 Factor X to the fourth power into X squared squared
00:24 The square root of any number (A) squared cancels out the square
00:28 Apply this formula to our exercise
00:31 This is the solution

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following exercise:

36x4= \sqrt{36x^4}=

2

Step-by-step solution

In order to simplify the given expression, apply the following three laws of exponents:

a. Definition of root as an exponent:

an=a1n \sqrt[n]{a}=a^{\frac{1}{n}}

b. Law of exponents for an exponent applied to terms in parentheses:

(ab)n=anbn (a\cdot b)^n=a^n\cdot b^n

c. Law of exponents for an exponent raised to an exponent:

(am)n=amn (a^m)^n=a^{m\cdot n}

Begin by converting the fourth root to an exponent using the law of exponents mentioned in a.:

36x4=(36x4)12= \sqrt{36x^4}= \\ \downarrow\\ (36x^4)^{\frac{1}{2}}=

We'll continue, using the law of exponents mentioned in b. and apply the exponent to each factor in the parentheses:

(36x4)12=3612(x4)12 (36x^4)^{\frac{1}{2}}= \\ 36^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}}

We'll continue once again, using the law of exponents mentioned in c. and perform the exponent applied to the term with an exponent in parentheses (the second factor in the multiplication):

3612(x4)12=3612x412=3612x2=36x2=6x2 36^{\frac{1}{2}}\cdot(x^4)^{{\frac{1}{2}}} = \\ 36^{\frac{1}{2}}\cdot x^{4\cdot\frac{1}{2}}=\\ 36^{\frac{1}{2}}\cdot x^{2}=\\ \sqrt{36}\cdot x^2=\\ \boxed{6x^2}

In the final steps, we first converted the power of one-half applied to the first factor in the multiplication back to the fourth root form, again, according to the definition of root as an exponent mentioned in a. (in the reverse direction) and then calculated the known fourth root of 36.

Therefore, the correct answer is answer d.

3

Final Answer

6x2 6x^2

Key Points to Remember

Essential concepts to master this topic
  • Root Conversion: Convert square root to fractional exponent: a=a12 \sqrt{a} = a^{\frac{1}{2}}
  • Exponent Distribution: Apply exponent to each factor: (ab)n=anbn (ab)^n = a^n \cdot b^n
  • Verification: Check by squaring the answer: (6x2)2=36x4 (6x^2)^2 = 36x^4

Common Mistakes

Avoid these frequent errors
  • Forgetting to apply the square root to both the coefficient and variable
    Don't just take the square root of 36 and ignore the x⁴ = 6x⁴! This gives a completely wrong form. The square root affects the entire expression under the radical. Always apply the square root operation to every factor inside.

Practice Quiz

Test your knowledge with interactive questions

Solve the following exercise:

\( \sqrt{\frac{2}{4}}= \)

FAQ

Everything you need to know about this question

Why does √(x⁴) equal x² and not x⁴?

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When you take the square root of x4 x^4 , you're asking "what times itself gives x4 x^4 ?" Since x2x2=x4 x^2 \cdot x^2 = x^4 , the answer is x2 x^2 .

How do I know when to use the exponent laws?

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Use exponent laws when you see roots, powers, or expressions in parentheses. Convert roots to fractional exponents first, then distribute exponents to each factor inside parentheses.

What if the coefficient under the square root isn't a perfect square?

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If the coefficient isn't a perfect square (like 20 instead of 36), look for perfect square factors. For example: 20=45=25 \sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5} .

Can I check my answer by plugging it back in?

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Yes! Square your final answer and see if you get the original expression. For 6x2 6x^2 : (6x2)2=36x4 (6x^2)^2 = 36x^4

Why do we assume x is positive in this problem?

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For basic square root problems, we typically assume variables represent non-negative values to avoid dealing with absolute value signs. This keeps the focus on the algebraic manipulation.

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