Extracting Square Roots: With fractions

Let's solve the given equation:

$\frac{x^2}{10}-10=0$

We'll start by getting rid of the fraction line in the left side of the given equation, we'll do this by multiplying both sides of the equation by the common denominator - which is the number 10, then we'll move the free number to one side, remembering that when moving a term between sides - its sign changes:

$\frac{x^2}{10}-\frac{10}{1}=0\hspace{8pt}\text{/}\cdot 10\\ \\ 1\cdot x^2-10\cdot10=0 \\ x^2-100=0\\ x^2=100$

From here, we'll solve simply, we'll perform on both sides the opposite operation to the square power operation applied to the unknown in the equation, which is the square root operation, using the laws of exponents:

a. Definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$and the two laws of exponents:

b. Law of exponents for power of a power:

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

Let's continue solving the equation:
$x^2=100\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ \sqrt{ x^2}=\pm\sqrt{ 100}\\ (x^2)^{\frac{1}{2}}=\pm10\\ x^{2\cdot\frac{1}{2}}=\pm10\\ \boxed{x=10,-10}$

In the first stage, we applied the square root to both sides of the equation, then we remembered the definition of root as a power (a.) on the left side, in the next stage - we applied the law of exponents for power of a power (b.) on the left side, and remembered that raising a number to the power of 1 doesn't change the number.

Additionally, we remembered that since an even power doesn't preserve the sign of the number it's applied to (will always give a positive result), taking an even root of both sides of the equation requires considering two possible cases - positive and negative (this is unlike taking a root of an odd order, which requires considering only one case that matches the sign of the number the root is applied to),

Let's summarize the solution of the equation:

$\frac{x^2}{10}-10=0 \hspace{8pt}\text{/}\cdot 10\\ x^2=100 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ \boxed{x=10,-10}$

Therefore, the correct answer is answer a.

Let's solve the given equation:

$\frac{x^5}{16}=2$

First, let's deal with the fact that we need to eliminate the denominator on the left side of the given equation, we'll do this by multiplying both sides of the equation by the common denominator - which is the number 16:

$\frac{x^5}{16}=2 \hspace{8pt}\text{/}\cdot 16\\ x^5=2\cdot 16\\ x^5=32$

From here, we'll solve simply, we'll perform on both sides the opposite operation to the fifth power that applies to the unknown in the equation, which is the fifth root operation, using the laws of exponents:

a. Definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$and two laws of exponents:

b. Law of exponents for power of a power:

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

Let's continue solving the equation:
$x^5=32 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \sqrt[5]{ x^5}=\sqrt[5]{ 32}\\ (x^5)^{\frac{1}{5}}=2\\ x^{5\cdot\frac{1}{5}}=2\\ \boxed{x=2}$

In the first stage, we applied the fifth root to both sides of the equation, then we remembered the definition of root as a power (a.) on the left side, in the next stage - we applied the law of exponents for power of a power (b.) on the left side, and remembered that raising a number to the power of 1 doesn't change the number.

Additionally, we remembered that since an odd-order power preserves the sign of the number it's applied to, taking an odd-order root requires consideration of only one case regarding the sign of the number being rooted (this is unlike taking an even-order root, which requires consideration of two possible cases - positive and negative),

Let's summarize the solution of the equation:

$\frac{x^5}{16}=2 \hspace{8pt}\text{/}\cdot 16\\ x^5=32 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \boxed{x=2}$

Therefore, the correct answer is answer B.

Let's solve the given equation:

$\frac{x^4}{16}=1$

Simply, we will perform on both sides the inverse operation of the fourth power that applies to the unknown in the equation, which is the fourth root operation. We'll use several laws of exponents:

a. Definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$and two laws of exponents:

b. Law of exponents for power of a power:

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

c. Law of exponents for power applied to a product in parentheses:

$\big(\frac{a}{b}\big)^n=\frac{a^n}{b^n}$

Let's proceed with solving the equation:

$\frac{x^4}{16}=1 \hspace{8pt}\text{/}\sqrt[4]{\hspace{6pt}}\\ \frac{\sqrt[4]{x^4}}{\sqrt[4]{16}}=\pm\sqrt[4]{ 1}\\ \frac{(x^4)^{\frac{1}{4}}}{2}=\pm1\\ \frac{x^{4\cdot\frac{1}{4}}}{2}=\pm1\\ \frac{x}{2}=\pm1\hspace{8pt}\text{/}\cdot 2\\ \boxed{x=2,-2}$

In the first stage, we applied the fourth root to both sides of the equation. Then we recalled the definition of root as a power (a.) on the left side and the law of exponents for power applied to a product in parentheses (c.). Additionally, we remembered that raising 1 to any power always yields 1. In the next stage, we applied the law of exponents for power of a power (b.) to the fraction's numerator on the left side, and remembered that raising a number to the power of 1 doesn't change the number. Finally, we eliminated the fraction on the left side by multiplying both sides of the equation by the common denominator - which is the number 2.

Furthermore, we remembered that since an even power doesn't preserve the sign of the number it's applied to (it will always give a positive result), taking an even root of both sides of the equation requires considering two possible cases - positive and negative (this is unlike taking a root of an odd order, which requires considering only one case that matches the sign of the number the root is applied to),

Therefore, the correct answer is answer c.

Let's solve the given equation:

$x^3=\frac{1}{8}$

Simply, we will perform on both sides the inverse operation of the cube power applied to the unknown in the equation, which is the cube root operation. We'll use several laws of exponents:

a. Definition of root as a power:

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

and two laws of exponents:

b. Law of exponents for power of power:

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

c. Law of exponents for power of parentheses:

$\big(\frac{a}{b}\big)^n=\frac{a^n}{b^n}$

Let's proceed with solving the equation:
$x^3=\frac{1}{8} \hspace{8pt}\text{/}\sqrt[3]{\hspace{6pt}}\\ \sqrt[3]{x^3}=\sqrt[3]{ \frac{1}{8}}\\ \sqrt[3]{x^3}= \frac{\sqrt[3]{1}}{\sqrt[3]{ 8}}\\ (x^3)^{\frac{1}{3}}=\frac{1}{\sqrt[3]{ 8}}\\ x^{3\cdot\frac{1}{3}}=\frac{1}{2}\\ \boxed{x=\frac{1}{2}}$

In the first step, we applied the cube root to both sides of the equation. Then we recalled the definition of root as a power (a.) on the left side and the law of exponents for power of parentheses (c.) on the right side. Additionally, we remembered that raising 1 to any power always yields 1. In the next step, we applied the law of exponents for power of power (b.), and remembered that raising a number to the power of 1 doesn't change the number,

Furthermore, we remembered that since an odd power preserves the sign of the number it's applied to, taking an odd root requires considering only one possible case which matches the sign of the number being rooted (this is unlike taking an even root, which requires considering two possible cases - positive and negative),

Therefore, the correct answer is answer d.

Let's solve the given equation:

$11x-\frac{1}{x}=10$

Let's start by getting rid of the fraction line in the left side of the given equation, we'll do this by multiplying both sides of the equation by the common denominator - which is the unknown $x$, while remembering that in the denominator (before multiplying by the common denominator - meaning in the given equation) cannot be zero since otherwise the fraction is undefined, therefore we must always define the domain accordingly, and require that the denominator is not zero(mentioned in the first line of the solution below), this step is a mandatory step when solving an equation:

$11x-\frac{1}{x}=10 \hspace{8pt}\text{/}\cdot x\hspace{8pt}\Leftrightarrow \boxed{ x\neq0}\\ x\cdot 11x-1\cdot 1 =x\cdot10 \\ 11x^2-1=10x\\ 11x^2-10x-1=0\\$In the final stage after we identified that we got a quadratic equation where the coefficient of the first-degree term (of the unknown) is not zero (meaning such a term exists in the equation), we moved all terms to one side,

From here, we'll solve using the quadratic formula, but before we do that in this problem-

Let's recall the quadratic formula:

The rule states that for a quadratic equation in the general form:

$ax^2+bx+c =0$

there are two solutions (or fewer) which we find using the formula:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Let's return now to the equation we got in the last stage:

$11x^2-10x-1=0$

Let's note that the coefficients from the general form we mentioned in the rule above:

$ax^2+bx+c$are:

$\begin{cases} a=11\\ b=-10\\ c=-1 \\ \end{cases}$

where we didn't forget to consider the coefficient together with its sign,

Therefore the solutions to the quadratic equation we got in the last stage are:

$x_{1,2}=\frac{-(-10)\pm\sqrt{(-10)^2-4\cdot11\cdot(-1)}}{2\cdot11}\\ x_{1,2}=\frac{10\pm\sqrt{100+44}}{22}\\ x_{1,2}=\frac{10\pm12}{22}\\ \downarrow\\ x_{1,2}=\frac{22}{22},\hspace{4pt}\frac{-2}{22}\\ \boxed{x=1,\hspace{4pt}-\frac{1}{11}}$

where in the final stage we simplified the fractions that were obtained as solutions,

Let's summarize then the solution of the equation:

$11x-\frac{1}{x}=10 \hspace{8pt}\text{/}\cdot x\hspace{8pt}\Leftrightarrow \boxed{ x\neq0}\\ 11x^2-1=10x\\ 11x^2-10x-1=0\\ \boxed{x=1,\hspace{4pt}-\frac{1}{11}}$

Note that both solutions we got for the unknown in the equation do not contradict the domain that was specified and therefore both are valid.

Therefore the correct answer is answer B.