Extracting Square Roots: With fractions

Let's solve the given equation:

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

Begin by eliminating the fraction line on the left side of the given equation. It is possible to achieve 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, 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. We then applied 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, given that 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),

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:

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

Begin by eliminating the denominator on the left side of the given equation, we can achieve 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, proceed to 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. Applying the definition of root as a power (a.) on the left side. In the next stage - we applied the law of exponents for a power of a power (b.) on the left side. Remember that raising a number to the power of 1 doesn't change the number.

Additionally, 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:

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

Perform the inverse operation of the fourth power (that applies to the unknown) on both sides of the equation. This is the fourth root operation. We will need to apply 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:

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

Proceed to solve 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. We then applied 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.). 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, due to the fact that an even power doesn't maintain 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:

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

Begin by eliminating the fraction line on the left side of the given equation. We can achieve this by multiplying both sides of the equation by the common denominator - which is the unknown $x$. Note that the denominator (before multiplying by the common denominator - meaning in the given equation) cannot be zero since the fraction would be undefined. Therefore we must always define the domain accordingly, and the denominator should not be 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 obtaining 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 proceed to solve the expression using the quadratic formula.

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 that we obtained in the last stage:

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

Note the coefficients from the general form that we mentioned in the rule above:

$ax^2+bx+c$are:

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

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

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

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

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 obtained 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.

Let's solve the given equation:

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

We will perform the inverse operation of the cube power (applied to the unknown) on both sides of the equation. This is the cube root operation. We will need to apply 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 to solve 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. Apply the definition of root as a power (a.) on the left side as well as the law of exponents for power of parentheses (c.) on the right side. Note that raising 1 to any power always yields 1. In the next step, we applied the law of exponents for power of power (b.) Remember that raising a number to the power of 1 doesn't change the number,

Furthermore, given that 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.