Solution by extracting a root

Solving a quadratic equation with one variable X2c=0X^2-c=0 (where b=0b=0) by calculating the square root:

First step

Moving terms and isolating X2X^2.

Second stage

Take the square root of both sides. Don't forget to insert ±\pm before the square root of the free term.

Third stage

Writing solutions in an organized manner or writing "no solution" in case of a root of a negative number.

Suggested Topics to Practice in Advance

  1. The quadratic equation
  2. Methods for Solving a Quadratic Function
  3. Squared Trinomial

Practice Extracting Square Roots

Examples with solutions for Extracting Square Roots

Exercise #1

Solve the following exercise:

2x28=x2+4 2x^2-8=x^2+4

Video Solution

Step-by-Step Solution

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

2x2x284=0 2x^2-x^2-8-4=0

We simplify :

x212=0 x^2-12=0

We use the shortcut multiplication formula to solve:

x2(12)2=0 x^2-(\sqrt{12})^2=0

(x12)(x+12)=0 (x-\sqrt{12})(x+\sqrt{12})=0

x=±12 x=\pm\sqrt{12}

Answer

±12 ±\sqrt{12}

Exercise #2

Solve the following equation:


2x28=x2+4 2x^2-8=x^2+4

Video Solution

Step-by-Step Solution

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

  • Step 1: Rearrange the equation.

We start with the given equation:

2x28=x2+42x^2 - 8 = x^2 + 4

Subtract x2+4x^2 + 4 from both sides to get:

2x28x24=02x^2 - 8 - x^2 - 4 = 0

  • Step 2: Simplify the equation.

Combine the like terms:

(2x2x2)84=0(2x^2 - x^2) - 8 - 4 = 0

This simplifies to:

x212=0x^2 - 12 = 0

  • Step 3: Solve for xx.

Add 12 to both sides:

x2=12x^2 = 12

Now take the square root of both sides:

x=±12x = \pm \sqrt{12}

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

Answer

±12 ±\sqrt{12}

Exercise #3

Solve the following equation:

x216=x+4 x^2-16=x+4

Video Solution

Step-by-Step Solution

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

Answer

5

Exercise #4

Solve the following equation:

x236=6x36 x^2-36=6x-36

Video Solution

Step-by-Step Solution

To solve the equation x236=6x36 x^2 - 36 = 6x - 36 , follow these steps:

  • Simplify the equation by subtracting 6x and 36 from both sides to get: x26x=0 x^2 - 6x = 0 .
  • Notice that the equation can be factored as it has a common factor: x(x6)=0 x(x - 6) = 0 .
  • According to the zero product property, set each factor equal to zero: x=0 x = 0 or x6=0 x - 6 = 0 .
  • Solve each resulting equation: x=0 x = 0 and x=6 x = 6 .

Therefore, the solutions to the quadratic equation x236=6x36 x^2 - 36 = 6x - 36 are x=0 x = 0 or x=6 x = 6 .

Answer

0 or 6 0~or~6

Exercise #5

Solve the following:

x2+x23=x2+6 x^2+x^2-3=x^2+6

Video Solution

Step-by-Step Solution

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

  • Simplify the given equation.
  • Solve for x2 x^2 and find x x .

First, let's simplify the equation:

x2+x23=x2+6 x^2 + x^2 - 3 = x^2 + 6 .

Combine like terms on the left side:

2x23=x2+6 2x^2 - 3 = x^2 + 6 .

Subtract x2 x^2 from both sides to isolate one of the x x terms:

2x2x23=6 2x^2 - x^2 - 3 = 6 .

This simplifies to:

x23=6 x^2 - 3 = 6 .

Next, add 3 to both sides to solve for x2 x^2 :

x2=9 x^2 = 9 .

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

x=±9 x = \pm\sqrt{9} .

This results in:

x=±3 x = \pm3 .

Therefore, the solution to the problem is ±3\pm3.

Answer

±3

Exercise #6

Solve the following exercise

x3x+7=2x2+9 x\cdot3\cdot x+7=2x^2+9

Video Solution

Step-by-Step Solution

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: x3x+7=3x2+7 x \cdot 3 \cdot x + 7 = 3x^2 + 7 .

2. Set up the equation by balancing: 3x2+7=2x2+9 3x^2 + 7 = 2x^2 + 9 .

3. Rearrange the terms to form a quadratic equation: 3x22x2+79=0 3x^2 - 2x^2 + 7 - 9 = 0 .

This simplifies to: x22=0 x^2 - 2 = 0 .

4. Solve for x x :
By adding 2 to both sides, we have: x2=2 x^2 = 2 .
Take the square root of both sides: x=±2 x = \pm\sqrt{2} .

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

Answer

±2 ±\sqrt{2}

Exercise #7

Solve the following equation:

4x2+8+2x=x+12+x 4x^2+8+2x=x+12+x

Video Solution

Step-by-Step Solution

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:

4x2+8+2x=x+12+x 4x^2 + 8 + 2x = x + 12 + x

The right-hand side can be simplified:

x+x+12=2x+12 x + x + 12 = 2x + 12

This yields the equation:

4x2+8+2x=2x+12 4x^2 + 8 + 2x = 2x + 12

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

4x2+8+2x2x12=0 4x^2 + 8 + 2x - 2x - 12 = 0

Combining like terms gives:

4x24=0 4x^2 - 4 = 0

Step 3: Solve the resulting quadratic equation:

First, we add 4 to both sides:

4x2=4 4x^2 = 4

Next, divide both sides by 4:

x2=1 x^2 = 1

Now, apply the square root to both sides:

x=±1 x = \pm 1

Therefore, the solutions to the quadratic equation are

x=±1 x = \pm 1 .

The correct answer is choice 2: ±1.

Answer

±1

Exercise #8

Solve the following exercise:

x220=5 x^2-20=5

Video Solution

Step-by-Step Solution

To solve this quadratic equation x220=5 x^2 - 20 = 5 , we will follow these steps:

  • Step 1: First, add 20 to both sides of the equation to isolate the x2 x^2 term:

x220+20=5+20 x^2 - 20 + 20 = 5 + 20

This simplifies to:

x2=25 x^2 = 25

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

x=±25 x = \pm \sqrt{25}

x=±5 x = \pm 5

Therefore, the solutions to the equation are:

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

Thus, the correct answer choice is:

  • ±5 \pm 5 from the provided options.

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

Answer

±5

Exercise #9

Solve for X:

xx=49 x\cdot x=49

Video Solution

Step-by-Step Solution

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

x249=0 x^2-49=0

x272=0 x^2-7^2=0

We use the abbreviated multiplication formula:

(x7)(x+7)=0 (x-7)(x+7)=0

x=±7 x=\pm7

Answer

±7

Exercise #10

Solve the following problem:

x2x=0 x^2-x=0

Video Solution

Step-by-Step Solution

Shown below is the given equation:

x2x=0 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 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:

x2x=0x(x1)=0 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:

x=0 \boxed{x=0}

or:

x1=0x=1 x-1=0\\ \downarrow\\ \boxed{x=1}

Let's summarize then the solution to the equation:

x2x=0x(x1)=0x=0x=0x1=0x=1x=0,1 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.

Answer

x=0,1 x=0,1

Exercise #11

x4+2x2=0 x^4+2x^2=0

Video Solution

Step-by-Step Solution

To solve the equation x4+2x2=0x^4 + 2x^2 = 0, we will use the technique of factoring. Let's proceed step-by-step:

First, notice that both terms x4x^4 and 2x22x^2 have a common factor of x2x^2. We can factor x2x^2 out from the equation:

x2(x2+2)=0x^2(x^2 + 2) = 0

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

  • x2=0x^2 = 0 or
  • x2+2=0x^2 + 2 = 0

Solving the first case, x2=0x^2 = 0:

x=0x = 0

For the second case, x2+2=0x^2 + 2 = 0, we reach:

x2=2x^2 = -2

Since x2=2x^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=0x = 0.

The correct choice from the provided options is:

x=0 x=0

Answer

x=0 x=0

Exercise #12

Solve for x:

x281=0 x^2-81=0

Video Solution

Step-by-Step Solution

Let's solve the given equation:

x281=0 x^2-81=0 Note that we can factor the expression on the left side using the difference of squares formula:

(a+b)(ab)=a2b2 (\textcolor{red}{a}+\textcolor{blue}{b}) (\textcolor{red}{a}-\textcolor{blue}{b})=\textcolor{red}{a}^2-\textcolor{blue}{b}^2 We'll do this using the fact that:

81=92 81=9^2 Therefore, we'll represent the rightmost term as a squared term:

x281=0x292=0 x^2-81=0 \\ \downarrow\\ \textcolor{red}{x}^2-\textcolor{blue}{9}^2=0 If so, we can represent the expression on the left side in the above equation as a product of expressions:

x292=0(x+9)(x9)=0 \textcolor{red}{x}^2-\textcolor{blue}{9}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{9})(\textcolor{red}{x}-\textcolor{blue}{9})=0 From here we'll remember that the product of expressions equals 0 only if at least one of the multiplied expressions equals zero,

Therefore, we'll get two simple equations and solve them by isolating the variable in each:

x+9=0x=9 x+9=0\\ \boxed{x=-9} or:

x9=0x=9 x-9=0\\ \boxed{x=9}

Let's summarize the solution of the equation:

x281=0x292=0(x+9)(x9)=0x+9=0x=9x9=0x=9x=9,9 x^2-81=0 \\ \downarrow\\ \textcolor{red}{x}^2-\textcolor{blue}{9}^2=0 \\ \downarrow\\ (\textcolor{red}{x}+\textcolor{blue}{9})(\textcolor{red}{x}-\textcolor{blue}{9})=0 \\ x+9=0\rightarrow\boxed{x=-9}\\ x-9=0\rightarrow\boxed{x=9}\\ \downarrow\\ \boxed{x=9,-9} Therefore, the correct answer is answer B.

Answer

x=±9 x=\pm9

Exercise #13

Solve the following equation:

7x1014x9=0 7x^{10}-14x^9=0

Video Solution

Step-by-Step Solution

Shown below is the given equation:

7x1014x9=0 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 7x9 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:

7x1014x9=07x9(x2)=0 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:

7x9=0/:7x9=0/9x=0 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:

x2=0x=2 x-2=0 \\ \boxed{x=2}

Let's summarize the solution of the equation:

7x1014x9=07x9(x2)=07x9=0x=0x2=0x=2x=0,2 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.

Answer

x=2,x=0 x=2,x=0

Exercise #14

Solve the following problem:

x6+x5=0 x^6+x^5=0

Video Solution

Step-by-Step Solution

Shown below is the given equation:

x6+x5=0 x^6+x^5=0

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

The largest common factor for the numbers and variables in this case is x5 x^5 given that the fifth power is the lowest power in the equation and therefore is included both in the term with the sixth power and in the term with the fifth power. Any power higher than this is not included in the term with the fifth 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. Proceed with the factoring of the expression:

x6+x5=0x5(x+1)=0 x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0

Let's continue 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, due to the fact 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:

x5=0/5x=0 x^5=0 \hspace{8pt}\text{/}\sqrt[5]{\hspace{6pt}}\\ \boxed{x=0} (in this case taking the odd root of the right side of the equation will yield one possibility)

or:

x+1=0x=1 x+1=0\\ \boxed{x=-1}

Let's summarize the solution of the equation:

x6+x5=0x5(x+1)=0x5=0x=0x+1=0x=1x=0,1 x^6+x^5=0 \\ \downarrow\\ x^5(x+1)=0 \\ \downarrow\\ x^5=0 \rightarrow\boxed{ x=0}\\ x+1=0 \rightarrow \boxed{x=-1}\\ \downarrow\\ \boxed{x=0,-1}

Therefore the correct answer is answer A.

Answer

x=1,x=0 x=-1,x=0

Exercise #15

Solve the following problem:

3x2+9x=0 3x^2+9x=0

Video Solution

Step-by-Step Solution

Shown below is the given problem:

3x2+9x=0 3x^2+9x=0

First, note that in the left side we are able to factor the expression by using a common factor. The largest common factor for the numbers and letters in this case is 3x 3x due to the fact that the first power is the lowest power in the equation and therefore 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. Therefore this is the term with the highest power that can be factored out as a common factor from all terms for the letters,

For the numbers, note that 9 is a multiple of 3, therefore it is the largest common factor for the numbers in both terms of the expression,

Let's continue to factor the expression:

3x2+9x=03x(x+3)=0 3x^2+9x=0 \\ \downarrow\\ 3x(x+3)=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:

3x=0/:3x=0 3x=0 \hspace{8pt}\text{/}:3\\ \boxed{x=0}

In solving the above equation, we divided both sides of the equation by the term with the variable,

Or:

x+3=0x=3 x+3=0 \\ \boxed{x=-3}

Let's summarize the solution of the equation:

3x2+9x=03x(x+3)=03x=0x=0x+3=0x=3x=0,3 3x^2+9x=0 \\ \downarrow\\ 3x(x+3)=0 \\ \downarrow\\ 3x=0 \rightarrow\boxed{ x=0}\\ x+3=0\rightarrow \boxed{x=-3}\\ \downarrow\\ \boxed{x=0,-3}

Therefore the correct answer is answer C.

Answer

x=0,x=3 x=0,x=-3

Topics learned in later sections

  1. Completing the square in a quadratic equation