Extracting Square Roots - Examples, Exercises and Solutions

First, we move the terms to one side equal to 0.

$2x^2-x^2-8-4=0$

We simplify :

$x^2-12=0$

We use the shortcut multiplication formula to solve:

$x^2-(\sqrt{12})^2=0$

$(x-\sqrt{12})(x+\sqrt{12})=0$

$x=\pm\sqrt{12}$

To solve this quadratic equation $x^2 - 20 = 5$, we will follow these steps:

• Step 1: First, add 20 to both sides of the equation to isolate the $x^2$ term:

$x^2 - 20 + 20 = 5 + 20$

This simplifies to:

$x^2 = 25$

• Step 2: Next, take the square root of both sides to solve for $x$:

$x = \pm \sqrt{25}$

$x = \pm 5$

Therefore, the solutions to the equation are:

$x = 5$ and $x = -5$

Thus, the correct answer choice is:

• $\pm 5$ from the provided options.

The correct solution to the problem is $\pm 5$.

We first rearrange and then set the equations to equal zero.

$x^2-49=0$

$x^2-7^2=0$

We use the abbreviated multiplication formula:

$(x-7)(x+7)=0$

$x=\pm7$

To solve this problem, we'll follow these steps:

• Simplify the given equation.
• Solve for $x^2$ and find $x$.

First, let's simplify the equation:

$x^2 + x^2 - 3 = x^2 + 6$.

Combine like terms on the left side:

$2x^2 - 3 = x^2 + 6$.

Subtract $x^2$ from both sides to isolate one of the $x$ terms:

$2x^2 - x^2 - 3 = 6$.

This simplifies to:

$x^2 - 3 = 6$.

Next, add 3 to both sides to solve for $x^2$:

$x^2 = 9$.

To find $x$, take the square root of both sides:

$x = \pm\sqrt{9}$.

This results in:

$x = \pm3$.

Therefore, the solution to the problem is $\pm3$.

To solve this problem, we'll follow these steps:

• Step 1: Simplify both sides of the equation.
• Step 2: Rearrange terms to form a quadratic equation.
• Step 3: Solve the quadratic equation using an appropriate method, such as factoring or applying the quadratic formula.

Now, let's work through each step:

Step 1: Simplify both sides of the equation:

The right-hand side can be simplified:

This yields the equation:

Step 2: Rearrange the terms to form a quadratic equation. Subtract $2x + 12$ from both sides:

Combining like terms gives:

Step 3: Solve the resulting quadratic equation:

First, we add 4 to both sides:

Next, divide both sides by 4:

Now, apply the square root to both sides:

Therefore, the solutions to the quadratic equation are

$x = \pm 1$.

The correct answer is choice 2: ±1.

Let's solve the equation step-by-step:

• Step 1: Rearrange the equation.

We start with the given equation:

$2x^2 - 8 = x^2 + 4$

Subtract $x^2 + 4$ from both sides to get:

$2x^2 - 8 - x^2 - 4 = 0$

• Step 2: Simplify the equation.

Combine the like terms:

$(2x^2 - x^2) - 8 - 4 = 0$

This simplifies to:

$x^2 - 12 = 0$

• Step 3: Solve for $x$.

Add 12 to both sides:

$x^2 = 12$

Now take the square root of both sides:

$x = \pm \sqrt{12}$

Given the choices, the correct answer is $\pm \sqrt{12}$.

Please note that the equation can be arranged differently:

x²-16 = x +4

x² - 4² = x +4

We will first factor a trinomial for the section on the left

(x-4)(x+4) = x+4

We will then divide everything by x+4

(x-4)(x+4) / x+4 = x+4 / x+4

x-4 = 1

x = 5

To solve this problem, we'll follow these steps:

• Simplify the given expression.
• Rearrange into standard quadratic form.
• Solve using applicable method.

Let's begin the process:
1. Simplify the left-hand side: $x \cdot 3 \cdot x + 7 = 3x^2 + 7$.

2. Set up the equation by balancing: $3x^2 + 7 = 2x^2 + 9$.

3. Rearrange the terms to form a quadratic equation: $3x^2 - 2x^2 + 7 - 9 = 0$.

This simplifies to: $x^2 - 2 = 0$.

4. Solve for $x$:
By adding 2 to both sides, we have: $x^2 = 2$.
Take the square root of both sides: $x = \pm\sqrt{2}$.

Therefore, the solution to the problem is $\pm\sqrt{2}$.

To solve the equation $x^2 - 36 = 6x - 36$, follow these steps:

• Simplify the equation by subtracting 6x and 36 from both sides to get: $x^2 - 6x = 0$.
• Notice that the equation can be factored as it has a common factor: $x(x - 6) = 0$.
• According to the zero product property, set each factor equal to zero: $x = 0$ or $x - 6 = 0$.
• Solve each resulting equation: $x = 0$ and $x = 6$.

Therefore, the solutions to the quadratic equation $x^2 - 36 = 6x - 36$ are $x = 0$ or $x = 6$.

To solve the equation $x^4 + 2x^2 = 0$, we will use the technique of factoring. Let's proceed step-by-step:

First, notice that both terms $x^4$ and $2x^2$ have a common factor of $x^2$. We can factor $x^2$ out from the equation:

$x^2(x^2 + 2) = 0$

Now, to solve for $x$, we apply the Zero Product Property, which gives us that at least one of the factors must be zero:

• $x^2 = 0$ or
• $x^2 + 2 = 0$

Solving the first case, $x^2 = 0$:

$x = 0$

For the second case, $x^2 + 2 = 0$, we reach:

$x^2 = -2$

Since $x^2 = -2$ has no real solutions (squares of real numbers are non-negative), we can conclude that this equation doesn't provide additional real solutions.

Therefore, the only real solution to the given equation is $x = 0$.

The correct choice from the provided options is:

$x=0$

Shown below is the given equation:

$x^2-x=0$

First note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is $x$and this is due to the fact that the first power is the lowest power in the equation. Therefore it is included both in the term with the second power and in the term with the first power. Any power higher than this is not included in the term with the first power, which is the lowest. Hence this is the term with the highest power that can be factored out as a common factor from all terms in the expression. Continue to factor the expression:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0$

Proceed to the left side of the equation that we obtained in the last step. There is a multiplication of algebraic expressions and on the right side the number 0. Therefore given that the only way to obtain 0 from a multiplication is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$\boxed{x=0}$

or:

$x-1=0\\ \downarrow\\ \boxed{x=1}$

Let's summarize then the solution to the equation:

$x^2-x=0 \\ \downarrow\\ x(x-1)=0 \\ \downarrow\\ x=0 \rightarrow\boxed{ x=0}\\ x-1=0 \rightarrow \boxed{x=1}\\ \downarrow\\ \boxed{x=0,1}$

Therefore the correct answer is answer B.

To solve this problem, we'll follow these steps:

• Step 1: Recognize the structure of the equation as a difference of squares.
• Step 2: Factor the equation.
• Step 3: Use the zero-product property to find the values of $x$.

Now, let's work through each step:
Step 1: The equation given is $x^2 - 16 = 0$. This equation is a type of difference of squares, as it can be expressed in the form $a^2 - b^2 = (a - b)(a + b)$.
Step 2: Recognizing $16$ as $4^2$, we can factor the equation: $x^2 - 16 = (x - 4)(x + 4) = 0$.
Step 3: According to the zero-product property, if the product of two expressions is zero, then at least one of the expressions must be zero. Therefore, we set each factor equal to zero:
$x - 4 = 0$ or $x + 4 = 0$.
Solving these simple linear equations gives $x = 4$ and $x = -4$.

Therefore, the solutions to the equation are $x_1 = -4$ and $x_2 = 4$.

To solve the equation $x^2 - 25 = 0$, follow these steps:

• Step 1: Isolate the square term: Begin by rewriting the equation to isolate $x^2$:

$x^2 - 25 = 0$
Add $25$ to both sides to obtain:
$x^2 = 25$

• Step 2: Extract the square roots: To solve $x^2 = 25$, take the square root of both sides. Remember to consider both the positive and negative roots:

$x = \pm \sqrt{25}$

• Step 3: Simplify: Calculate the square root of 25:

$\sqrt{25} = 5$

Therefore, the solutions are:
$x = 5$
and
$x = -5$

Thus, the solutions to the equation $x^2 - 25 = 0$ are $x_1 = 5$ and $x_2 = -5$.

Verifying with the provided choices, the correct choice matches the solution $x_1 = 5, x_2 = -5$.

Therefore, the solution to the problem is $x_1 = 5, x_2 = -5$.

Solve the given equation:

$7x^3-x^2=0$

Note that on the left side we are able to factor the expression using a common factor. The largest common factor for the numbers and letters in this case is $x^2$since the second power is the lowest power in the equation and therefore is included both in the term with the third power and in the term with the second power. Any power higher than this is not included in the term with the second power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms in the expression.

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0$

Note that the left side of the equation that we obtained in the last step is a multiplication of algebraic expressions and on the right side the number 0.

Therefore, given that the only way to obtain 0 from a multiplication operation is to multiply by 0. Hence at least one of the expressions in the multiplication on the left side must equal zero,

meaning:

$x^2=0 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ x=\pm0\\ \boxed{x=0}$(in this case taking the even root of the right side of the equation will indeed yield two possibilities, positive and negative. However since we're dealing with zero, we'll get only one possibility)

or:

$7x-1=0$Let's solve this equation in order to obtain the additional solutions (if they exist) to the given equation:

We obtained a simple first-degree equation which we'll solve by isolating the unknown on one side, we'll do this by moving terms and then dividing both sides of the equation by the coefficient of the unknown:

$7x-1=0 \\ 7x=1\hspace{8pt}\text{/}:7\\ \boxed{x=\frac{1}{7}}$

Let's summarize the solution of the equation:

$7x^3-x^2=0 \\ \downarrow\\ x^2(7x-1)=0\\ \downarrow\\ x^2=0 \rightarrow\boxed{ x=0}\\ 7x-1=0\rightarrow \boxed{x=\frac{1}{7}}\\ \downarrow\\ \boxed{x=0,\frac{1}{7}}$

Therefore the correct answer is answer C.

Shown below is the given equation:

$7x^{10}-14x^9=0$

First, note that on the left side we are able to factor the expression using a common factor.

The largest common factor for the numbers and variables in this case is $7x^9$given that the ninth power is the lowest power in the equation and therefore is included in both the term with the tenth power and the term with the ninth power. Any power higher than this is not included in the term with the ninth power, which is the lowest, and therefore this is the term with the highest power that can be factored out as a common factor from all terms for the variables,

For the numbers, note that 14 is a multiple of 7, therefore 7 is the largest common factor for the numbers in both terms of the expression,

Let's continue and perform the factoring:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0$

On the left side of the equation that we obtained in the last step there is a multiplication of algebraic expressions and on the right side the number 0, therefore, since the only way to obtain a result of 0 from a multiplication operation is to multiply by 0, at least one of the expressions in the multiplication on the left side must equal zero,

Meaning:

$7x^9=0 \hspace{8pt}\text{/}:7\\ x^9=0 \hspace{8pt}\text{/}\sqrt[9]{\hspace{6pt}}\\ \boxed{x=0}$

In solving the equation above, we first divided both sides of the equation by the term with the variable, and then we proceeded to extract a ninth root from both sides of the equation.

(In this case, extracting an odd root from the right side of the equation yielded one possibility)

Or:

$x-2=0 \\ \boxed{x=2}$

Let's summarize the solution of the equation:

$7x^{10}-14x^9=0 \\ \downarrow\\ 7x^9(x-2)=0\\ \downarrow\\ 7x^9=0 \rightarrow\boxed{ x=0}\\ x-2=0\rightarrow \boxed{x=2}\\ \downarrow\\ \boxed{x=0,2}$

Therefore, the correct answer is answer A.