Square Root Quotient Property: Using variables

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.

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 root in the fraction's numerator in the problem:

$\frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}$

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 fraction's numerator in the expression we got in the last step:

$\frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}\cdot(x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}$

$$where in the first stage we applied the above-mentioned law of exponents noted 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,

Let's simplify the expression we got:

$\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x}=\frac{\sqrt{49}x^{\frac{2}{2}}}{x}=\frac{7x^1}{x}$

where in the first stage we converted back the fraction's power 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, but 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 we got, then we simplified the resulting fraction in that exponent for that factor.

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

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

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

$\frac{\sqrt{49x^2}}{x}=\frac{(49x^2)^{\frac{1}{2}}}{x}=\frac{49^{\frac{1}{2}}x^{2\cdot\frac{1}{2}}}{x} =7$

Therefore the correct answer is answer c.

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 root in the numerator of the fraction in the problem:

$\frac{\sqrt{x^4}}{x}=\frac{(x^4)^{\frac{1}{2}}}{x}$

Now let's remember the power law for power of power:

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

Let's apply this law to the numerator of the fraction in the expression we got in the last step:

$\frac{(x^4)^{\frac{1}{2}}}{x}=\frac{x^{4\cdot\frac{1}{2}}}{x}=\frac{x^\frac{4}{2}}{x}$

where in the first step we applied the above power law and in the second step we performed the multiplication in the power exponent of the numerator term,

Let's continue and simplify the expression we got, first we'll reduce the fraction with the power exponent in the numerator term and then we'll use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$

to simplify the fraction in the complete expression:

$\frac{x^\frac{4}{2}}{x}=\frac{x^2}{x}=x^{2-1}=x$

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

$$$\frac{\sqrt{x^4}}{x}=\frac{(x^4)^{\frac{1}{2}}}{x}=\frac{x^2}{x}=x$

Therefore the correct answer is answer A.