Simplify sqrt(25x²)/sqrt(x²): Step-by-Step Radical Reduction

Let's use the definition of root as a power:

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

when we remember that in a square root (also called "root to the power of 2") we don't write the root's power and:

$n=2$

meaning:

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

Let's return to the problem and convert using the root definition we mentioned above the roots in the problem:

$\frac{\sqrt{25x^2}}{\sqrt{x^2}}=\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}}$

Now let's recall two laws of exponents:

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

$(a\cdot b)^n=a^n\cdot b^n$

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

$(a^m)^n=a^{m\cdot n}$

Let's apply these laws to the numerator and denominator of the fraction in the expression we got in the last step:

$\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}}=\frac{25^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}}{(x^2)^{\frac{1}{2}}}=\frac{25^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x^{2\cdot\frac{1}{2}}}$

$$where in the first stage we applied the above law of exponents mentioned in a' and applied the power to both factors of the product in parentheses in the fraction's numerator, we did this carefully using parentheses since 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 in the fraction's numerator and similarly to the factor in the fraction's denominator,

Let's simplify the expression we got:

$\frac{25^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x^{2\cdot\frac{1}{2}}}=\frac{\sqrt{25}x^{\frac{2}{2}}}{x^{\frac{2}{2}}}=\frac{5x^1}{x^1}$

where in the first stage we converted back the fraction's power to a root, for the first factor in the product, this was done using the definition of root as a power mentioned at the beginning of the solution, but in the opposite direction, then we calculated the numerical value of the root,

Additionally - we calculated the product of the power of the second factor in the product in the fraction's numerator in the expression we got and similarly for the factor in the fraction's denominator, then we simplified the resulting fraction for that factor.

Let's complete the calculation and simplify the resulting fraction:

$\frac{5x^1}{x^1} =\frac{5\not{x}}{\not{x}}=5$

Let's summarize the solution steps so far, we got that:

$\frac{\sqrt{25x^2}}{\sqrt{x^2}}=\frac{(25x^2)^{\frac{1}{2}}}{(x^{2)^{\frac{1}{2}}}} =\frac{5x^1}{x^1} =5$

Therefore the correct answer is answer c.