Simplify the Radical Expression: √(49x²)/x

Radical Simplification with Square Roots

Solve the following exercise:

49x2x= \frac{\sqrt{49x^2}}{x}=

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:07 Let's solve this problem.
00:10 When we have a root of multiplied terms, like A times B.
00:14 We can split it into the root of A, times the root of B.
00:19 We'll use this idea in our exercise today.
00:28 The square root of any number A, that's squared, cancels the square.
00:33 We'll apply this to cancel out squares.
00:37 Now, let's simplify what we have.
00:43 Break 49 down into 7 squared.
00:47 Remember, the root cancels the square.
00:50 And that gives us our solution!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

Solve the following exercise:

49x2x= \frac{\sqrt{49x^2}}{x}=

2

Step-by-step solution

Express the following root as a power:

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

Remember that in a square root (also called "root to the power of 2") we don't write the root's power as shown

n=2 n=2

Meaning:

a=a2=a12 \sqrt{a}=\sqrt[2]{a}=a^{\frac{1}{2}}

Let's return to the problem and use the root definition that we mentioned above to convert the root in the fraction's numerator:

49x2x=(49x2)12x \frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}

Remember the two following laws of exponents:

a. The law of exponents for a power applied to a product inside of parentheses:

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

b. The law of exponents for a power of a power:

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

Let's apply these laws to the fraction's numerator in the expression that we obtained in the last step:

(49x2)12x=4912(x2)12x=4912x212x \frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}

In the first stage we applied the above-mentioned law of exponents noted in a' and then proceeded to applythe power to both factors of the product (in parentheses) in the fraction's numerator. We we careful to use parentheses given that one of the factors in the parentheses is already raised to a power.

In the second stage we applied the second law of exponents mentioned in b' to the second factor in the product,

Let's simplify the expression that we obtained:

4912x212x=49x22x=7x1x \frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}=\frac{\sqrt{49}x^{\frac{2}{2}}}{x}=\frac{7x^1}{x}

In the first stage we converted the fraction's power back to a root, for the first factor in the product, using the definition of root as a power mentioned at the beginning of the solution ( in the opposite direction)

Additionally- we calculated the product in the exponent of the second factor in the product in the fraction's numerator in the expression that we obtained. We then we simplified the resulting fraction.

Finish the calculation and proceed to simplify the resulting fraction:

7x1x=7=7 \frac{7x^1}{x}=\frac{7\not{x}}{\not{x}}=7

Let's summarize the various steps of the solution that we obtained thus far, as shown below:

49x2x=(49x2)12x=4912x212x=7 \frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x} =7

Therefore the correct answer is answer c.

3

Final Answer

7 7

Key Points to Remember

Essential concepts to master this topic
  • Rule: Simplify radicals by extracting perfect square factors first
  • Technique: Convert 49x2=7x \sqrt{49x^2} = 7|x| since 49 = 7²
  • Check: Substitute back: 7xx=7 \frac{7|x|}{x} = 7 when x > 0 ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting absolute value when simplifying square roots
    Don't assume x2=x \sqrt{x^2} = x always! This gives wrong answers when x is negative. x2=x \sqrt{x^2} = |x| because square roots are always non-negative. Always consider the domain and use absolute values when needed.

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 49x2 \sqrt{49x^2} equal 7x 7|x| and not just 7x 7x ?

+

Because square roots always give non-negative results! When x is negative, x2 x^2 is still positive, so x2=x \sqrt{x^2} = |x| . For example, if x = -3, then (3)2=9=3=3 \sqrt{(-3)^2} = \sqrt{9} = 3 = |-3| .

How do I know when I can cancel the x terms?

+

You can cancel when x ≠ 0 and both terms have the same sign. Since 7xx \frac{7|x|}{x} assumes x in the denominator, we need x > 0, making |x| = x, so 7xx=7 \frac{7x}{x} = 7 .

What if x could be negative in this problem?

+

If x < 0, then x=x |x| = -x , so 7xx=7(x)x=7 \frac{7|x|}{x} = \frac{7(-x)}{x} = -7 . But since we're dividing by x, the domain restriction x ≠ 0 applies, and the context usually assumes x > 0.

Can I solve this without using absolute value notation?

+

Yes, if you assume x > 0 from the start! Many textbooks make this assumption for simplicity. When x > 0, then x2=x \sqrt{x^2} = x directly, making 49x2x=7xx=7 \frac{\sqrt{49x^2}}{x} = \frac{7x}{x} = 7 .

Why is 7 the final answer and not 7x?

+

Because the x terms cancel out completely! We have 7xx \frac{7x}{x} (assuming x > 0), and since any number divided by itself equals 1, we get 7xx=71=7 7 \cdot \frac{x}{x} = 7 \cdot 1 = 7 .

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