how to solve this equation x²=16​

Step-by-step explanation:

Hey there!

To solve this type of question, you must take square of variable"X" to right side making it ±squareroot. As it is a quadratic equation it will have two values (±).

Given;

[tex] {x}^{2} = 16[/tex]

Taking (square) to right side. It becomes ± square root.

[tex]x = + - \sqrt{16} [/tex]

[tex]x = + - \sqrt{ {4}^{2} } [/tex]

Cancel square and square root.

[tex]x = + - 4[/tex]

Therefore, X = ±4

Hope it helps...

Answer:

There are two I results found which is x=4 and x=-4

Step-by-step explanation:

Step by Step Solution:

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Rearrange:

Rearrange the equation by subtracting what is to the right of the equal sign from both sides of the equation :

                    x^2-(16)=0  

Step by step solution :

STEP

1 :

Trying to factor as a Difference of Squares:

1.1      Factoring:  x2-16  

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

        A2 - AB + BA - B2 =

        A2 - AB + AB - B2 =

        A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 16 is the square of 4

Check :  x2  is the square of  x1  

Factorization is :       (x + 4)  •  (x - 4)  

Equation at the end of step

1 :

 (x + 4) • (x - 4)  = 0  

STEP

2 :

Theory - Roots of a product

2.1    A product of several terms equals zero.  

When a product of two or more terms equals zero, then at least one of the terms must be zero.  

We shall now solve each term = 0 separately  

In other words, we are going to solve as many equations as there are terms in the product  

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

2.2      Solve  :    x+4 = 0  

Subtract  4  from both sides of the equation :  

                     x = -4

Solving a Single Variable Equation:

2.3      Solve  :    x-4 = 0  

Add  4  to both sides of the equation :  

                     x = 4 and/or x=-4

[tex](x - 2) { }^{2} = 0[/tex]
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