If you extract the square root from both parts of the equation, you get an equation equal to the original one.

Consider the following equation:

x2 = 16

This is the simplest quadratic equation that has two roots: 4 and -4. We solved such an equation using the definition of a square root.

According to the definition of a square root, b is the square root of a if b2 = a and is denoted as b = √a.

Then in the case of x2 = 16, we can write that x = √16, whence x = ±4.

Now solve this quadratic equation by extracting the square root from both parts of the equation.

"Wrap" both parts of the equation x2 = 16 to the square root:

extraction of the square root from both parts of the equation Fig. 1

Now recall one of the properties of the square root, which states that the square root of the square of a number is equal to the modulus of that number

2 = a

Then on the left side of our equation we get the modulus of x, and on the right side we get the number 4

extraction of the square root from both parts of the equation Fig. 2

We obtained the simplest equation with a modulus. It has two roots: 4 and -4. Write down this solution as a set of equations:

extraction of the square root from both parts of the equation Fig. 4

Checking:

extraction of the square root from both parts of the equation Fig. 5

It is the arithmetic square root that must be extracted from the right side of the equation x2 = 16. Earlier we said that the square root has two values: positive and negative. That is:

extraction of the square root from both parts of the equation Fig. 9

But in this case we are interested in the non-negative value of 4 (which is called the arithmetic square root). Because if we extract the second root (negative -4), we get the equation |x|= -4, which has no solutions.


Example 2. Solve the equation 3x2 = 12

Solution

Divide both parts by 3

extraction of the square root from both parts of the equation Fig. 6

Extract the square root of both parts of the resulting equation:

extraction of the square root from both parts of the equation Fig. 7

We obtained the simplest equation with a module. Solve it by reducing it to a set:

extraction of the square root from both parts of the equation Fig. 8

Answer: 2 and -2.


Example 3. Solve the equation (x + 2)2 = 25

Solution

Extract the square root of both parts of the resulting equation:

extraction of the square root from both parts of the equation Fig. 9

Solve the resulting equation with a module:

extraction of the square root from both parts of the equation Fig. 10

Answer: 3 and -7.


Example 4. Solve the equation x2 - 10 = 39

Solution

Move -10 to the right side by changing the sign:

extraction of the square root from both parts of the equation Fig. 11

Extract the square root of both parts of the resulting equation:

extraction of the square root from both parts of the equation Fig. 12

Solve the resulting equation with a module:

extraction of the square root from both parts of the equation Fig. 13

Answer: 7 and -7.


Tasks for independent decision

Task 1. Solve the equation:
Solution:
Answer: 9 and −9.
Task 2. Solve the equation:
Solution:
Answer: 0.4 and −0.4.
Task 3. Solve the equation:
Solution:
Answer: 4 and −4.
Task 4. Solve the equation:
Solution:
Answer: and .
Task 5. Solve the equation:
Solution:
Answer: and .

Add comment

1. If you don't understand something, you can ask a question through the form below
2. If you find an error or inaccuracy, please describe it.
3. Positive feedback is welcome.


Security code
Refresh