Solution by extracting a root

🏆Practice extracting square roots

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.

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Test yourself on extracting square roots!

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Solve for X:

\( x\cdot x=49 \)

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Solution by extracting a root

Pay attention -
This solution method is suitable for quadratic equations where there is no bb, and Y=0Y=0, meaning quadratic equations that look like the following:
Y=x2+cY=x^2+c
For example:
X2+4=0X^2+4=0
In equations of this type, we won't need to use the quadratic formula or other long methods. Instead we can use a much more efficient and quicker method- by calculating the square root!

How do we solve it?

Let's learn through an example.
Let's solve the following equation:
x216=0x^2-16=0

First step:

Move the terms and isolate x2x^2 so it will be alone on one side (even without a coefficient).
Perform the following calculation:
x2=16x^2=16

Second stage:

Let's take the square root of both sides and remember to insert ±± due to the fact that when we take the square root of a free number there are 22 opposite solutions (one negative and one positive)
Let's solve the following:
x=±4x=±4

Third stage:

Let's write the answers that we obtained in an organized way.
Proceed as follows:
X1=4X_1=4
X2=4X_2=-4

Excellent! Now we'll continue practicing solving equations with one variable 11, increase the difficulty level and encounter different cases.

Exercise with coefficient greater than 11 for x2x^2:
3x29=03x^2-9=0

Solution:
Let's start by completely isolating x2x^2. We'll perform the following:
3x2=93x^2=9
x2=3x^2=3

Now, after we completely isolated x2x^2, we can take the square root of both sides. Let's not forget to place the ±\pm sign
Continue to solve:
x=±3x=±\sqrt3
Let's proceed to write the answers in an organized way:
X1=3X_1=\sqrt3
X2=3 X_2= -\sqrt3

Exercise requiring finding the square root of a negative number:
X2+4=0X^2+4=0

Solution:
To begin with, we'll move terms in order to isolate X2X^2
We'll perform the following:
x2=4x^2=-4
Now we need to find the square root as usual. But!! We cannot find the square root of a negative number!
This means that there is no number that when squared will give us a negative result.
Therefore when we try to find the square root:
X=(4)X=\sqrt{(-4)}
The answer will be - no solution!
Note! If you leave the answer like this
X=(4)X=\sqrt{(-4)}
it would be a mistake. You must write - no solution. The square root of a negative number cannot be found.

Another exercise:
12x2240=012x^2-240=0

Solution:
At first glance, this exercise might look slightly intimidating given that its numbers are relatively large. However, we can solve exercises like these easily using the method we've learned so far.
To begin with, we completely isolate x2x^2 .
We perform the following:
12X2=24012X^2=240
Divide both sides by 1212 as follows:
x2=20x^2=20
Now we'll take the square root and not forgetting to insert the ±\pm sign
We'll perform the following:
X=±20X=±\sqrt{20}
Write both answers in a concise way:
X1=20X_1=-\sqrt{20}

X2=20X_2=\sqrt{20}

Another exercise:
3x2+150=03x^2+150=0

Solution:
To begin with let's move terms to obtain the following:
3x2=1503x^2=-150
Let's proceed to divide both sides by 33 in order to completely isolate the X2X^2.
We obtain the following:
X2=50X^2=-50
At this point, we can observe that the equation has no solution due to the fact that it's impossible for any number squared (whether positive or negative) to give us a negative number like 50-50.
Nevertheless, let's continue to prove that there's no solution and try to take the square root.
We obtain the following:
x=±(50)x=±\sqrt{(-50)}
The answer will be - no solution since we cannot take the square root of a negative number.

Important note –
Sometimes students get confused and tend to think there is only 11 solution to the equation, the solution of the ++.
This is a serious mistake since we still need to find the square root of 5050- and not of 5050, therefore we won't obtain any solution when we need to find the square root of a negative number.

Exercise with a fractional result:
3X24=03X^2-4=0

Solution:
To begin with, we'll move terms to obtain the following:
3x2=43x^2=4
Now we'll divide both sides by 33 :
x2=43x^2=\frac43
Note, don't be alarmed by getting a fraction, we'll continue as usual. The next step is to take the square root. We obtain the following:
x=±43x=±\sqrt{\frac43}
We'll the proceed to write the results in an organized way:
X1=43X_1=\sqrt{\frac43}
X2=43X_2=-\sqrt{\frac43}

Important note - when a square root can be easily simplified, like for the number 16\sqrt{16} for example, you should not leave the answer like that and write 44.
However, if you cannot get a whole number square root, it is usually fine to leave the answer with the square root.

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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:

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 #3

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 #4

x2x=0 x^2-x=0

Video Solution

Step-by-Step Solution

The equation in the problem is:

x2x=0 x^2-x=0

First let's note that in the left side we can 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 because 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, 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, so we'll continue and perform the factoring:

x2x=0x(x1)=0 x^2-x=0 \\ \downarrow\\ x(x-1)=0

Let's continue and address the fact that in the left side of the equation we got 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 get 0 from 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 #5

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

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