Product Property of Square Roots: Using variables

In order to simplify the given expression, we will use the following three laws of exponents:

a. The definition of root as an exponent:

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

b. Law of exponents for power to a power:

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

Let's start with converting the square root to an exponent using the law of exponents mentioned in a:

$\sqrt{x^4}= \\ \downarrow\\ (x^4)^{\frac{1}{2}}=$Let's continue, using the law of exponents mentioned in b to perform the exponentiation of the term in parentheses:

$(x^4)^{{\frac{1}{2}}} = \\ x^{4\cdot\frac{1}{2}}=\\ \boxed{x^2}$Therefore, the correct answer is answer b.

In order to simplify the given expression, we will use two laws of exponents:

A. Definition of the root as an exponent:

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

B. Law of exponents for dividing powers with the same base:

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

Let's start with converting the root to an exponent using the law of exponents shown in A:

$\sqrt{9x}= \\ \downarrow\\ (9x)^{\frac{1}{2}}=$Next, we will use the law of exponents shown in B and apply the exponent to each of the factors in the numerator that are in parentheses:

$(9x)^{\frac{1}{2}}= \\ 9^{\frac{1}{2}}\cdot x^{{\frac{1}{2}}}=\\ \sqrt{9}\sqrt{x}=\\ \boxed{3\sqrt{x}}$$$In the last steps, we will multiply the half exponent by each of the factors in the numerator, returning to the root form, that is, according to the definition of the root as an exponent shown in A (in the opposite direction) and then we will calculate the known fourth root result of the number 9.

Therefore, the correct answer is answer D.

Given that:

$\frac{AB}{BC}=\sqrt{\frac{x}{2}}$

Given that AB equals X

We will substitute accordingly in the formula:

$\frac{x}{BC}=\frac{\sqrt{x}}{\sqrt{2}}$

$x\sqrt{2}=BC\sqrt{x}$

$\frac{x\sqrt{2}}{\sqrt{x}}=BC$

$\frac{\sqrt{x}\times\sqrt{x}\times\sqrt{2}}{\sqrt{x}}=BC$

$\sqrt{x}\times\sqrt{2}=BC$

Now let's focus on triangle ABC and use the Pythagorean theorem:

$AB^2+BC^2=AC^2$

Let's substitute the known values:

$x^2+(\sqrt{x}\times\sqrt{2})^2=m^2$

$x^2+x\times2=m^2$

We'll add 1 to both sides:

$x^2+2x+1=m^2+1$

$(x+1)^2=m^2+1$