A quadratic equation is an equation that has an x² and is equal to 0. On a graph, the lines will look like a v. This is because, when solved, this equation will have two answers. There are 3 ways to solve a quadratic equation, I’ll be showing how to use one of them, factoring.

Now, we have a couple of questions we must ask ourselves when factoring.

Is there anything in common? Remove it.

15x³ – 5x²

5x²(3x – 1)

Is it a difference of squares?

There is a difference of 3x and 1, but they are not both squares.

25x²-49 would be a difference of squares that could then be factored.

25x²-49

(5x – 7)(5x + 7)

If we are still left with factorable terms, and there are three terms left, is there a pattern? Is it easy to factor or hard?

The pattern we look for is ax² + bx + c

If a = 1, it is easy to factor and we just need two terms that have the product of c and the sum of b.

x² + 7x + 10

The factors of ten are : 1 x 10, 2 x 5

2 and 5 have the sum of 7

Both terms are positive so the factor is then (x + 2)(x + 5)

If a is a number higher than 1, we can use methods such as guess and test.

5x² + 9x + 4

Since the sum of 5 and 4 is 9, we can guess that the factor is

(5x + 4)(x + 1)

Because we distributed, it equals 5x² + 9x + 4

 

Now, we can use what we know of factoring terms to solve a quadratic equation.

A quadratic equation looks like

s² – 2s – 35 = 0

There is an x² term, and it is equal to 0.

 

To solve this, we can factor one side. Since there is no common factors, we can skip difference of squares (since there is 3 terms, not 2) and factor the pattern the easy way. -7 and 5 have the sum of -2 and -35.

(s – 7)(s + 5) = 0

s – 7 = 0

+ 7

s = 7

s + 5 = 0

-5

s = -5

s equals 7 or -5

When there is an equality sign (=) in an equation, we can further solve it.

For example: √x=10 is a radical equation

We can solve this equation by isolating the variable, x.

√x=10

To rid of the √, we can square both sides.

√x²=10²

x=100

Voila, we have found what x equals. To finish, since there was a variable under a √ sign, we need to show restrictions.

This is done by taking whatever is under the radical with an even index, and saying that x≥0.

Mind you, if the index of the radical wasn’t a multiple of 2, but an odd number like 3, x is an element of the real numbers and has no restrictions.

Then, we should always check to make sure our answer is correct.

√x=10

√100

10 = 10

 

Another example would be the equation √x+3 = 5, where x + 3 is all under the radical sign.

We do the same thing of squaring both sides

(√x+3)² = 5²

x+3 = 25

Isolate the variable by removing -3 from both sides

x = 22

Now, we can find the restriction

x+3≥0

Minus 3

x≥-3

X is greater than -3 so our restriction is correct and we can do a check of the equation

√x+3 = 5

√22+3

√25

5 = 5

This answer works.

 

Some things to keep in mind

Some answers may not work, they are called extraneous solutions and have no solutions. That is why it is important to check the answer.

When finding restrictions, if there is a need to divide a negative term from both sides, the sign switches. This means that if it is 7-6x≥0

We isolate the variable by first -7

-6x≥7

Since we then divide by a negative 6, the sign changes

x≤7⁄6