Solving Quadratic Equations

Algebra Unit – Part 1

( x – 2) ( x

+ 8 )

( x – 7) ( x – 3 )

( x + 4) ( x + 6 )

( x + 1) ( x - 5 )

Why did you learn to factor a trinomial?

You learned to factor a trinomial into TWO binomials in order to use those answers.

In order to use those answers you have to set each binomial = to zero (0).

( X – 3 ) ( x + 5 ) = 0 If you multiply two binomials and the

value of one of them is zero, then the whole product is zero.

Setting the binomial = to Zero The product ( x – 3 ) ( x + 5 ) = 0 If either binomial = zero then the whole

things is zero. What number would make the 1st binomial = 0?▪ 3

What number would make the 2nd binomial = 0?▪ -5

That means if x = 3 or x = -5, the whole problem is 0. Therefore, our answer is x = 3 or x = -5.

Sample Problems

In the middle of your sheet of notes fill in the answers as we go, by setting each binomial = to zero.

( x – 2 ) ( x + 8) = 0 X = 2 or x = -8

( x – 7 ) ( x – 3 ) = 0 X = 7 or x = 3

( x + 4 ) ( x + 6 ) = 0 X = -4 or x = -6

( x + 1 ) ( x – 5 ) = 0 X = -1 or x = 5

Solving Basic Square Root Problems

The easiest type of quadratics to solve is basic square root problems.

They come in two forms: x2= 64 and x2 – 36 = 0

To solve the first one, all you do is take the square root of the number. X = √64 = 8 and -8

To solve the second one, you have to add 36 to both sides and then take the square root. x2 – 36 = 0 +36 +36 x2 = 36X = √36 = 6 and -6

Sample Problems

x2= 81 X = √81 = 9 and -9

x2= 9 X = √9 = 3 and -3

x2 = 16 X = √16 = 4 and -4

x2 – 25 = 0 +25 +25 x2 = 25X = √25 = 5 and -5 x2 – 49 = 0 +49 +49 x2 = 49X = √49 = 7 and -7 x2 – 4 = 0 +4 +4 x2 = 4X = √4 = 2 and -2

At the end of your sheet of notes fill in the answers as we go, by setting each binomial = to zero.

Steps to solving regular quadratics

1. Set the trinomial = to zero.

2. Factor the trinomial into the product of binomials

3. Set each binomial = to zero

4. Solve for x

Example Solve: x2+ 5x - 24

= 0 Only way to get a negative at the end

is multiply 1 positive & 1 negative, looking at middle number the bigger number needs to be positive.

( x + 8 ) ( x - 3) = 0

( x + 8 ) = 0 or ( x – 3) = 0

X = -8 or x = 3

Sample regular quadratic problems

x2+ 12x + 32 = 0 ( x + 8 ) ( x + 4) =

0 ( x + 8 ) = 0 or ( x

+ 4) = 0 X = -8 or x = -4

X2- 8x + 15 = 0 ( x - 5 ) ( x - 3) = 0 ( x - 5 ) = 0 or ( x -

3) = 0 X = 5 or x = 3

In the middle of your sheet of notes fill in the answers as we go, by creating binomials and then setting each binomial = to zero. X2- 8x - 9 = 0

( x - 9 ) ( x + 1) = 0 ( x - 9 ) = 0 or ( x +

1) = 0 X = 9 or x = -1

Adding one more step

Again the first thing you need to do is set the trinomial = to zero. Therefore you may need to add or subtract in order to do so.

Example x2 + 13x = -30 +30 + 30

x2+ 13x + 30 = 0 ( x + 10 ) ( x + 3) = 0 ( x + 10 ) = 0 or ( x + 3) = 0 X = -10 or x = -3

Sample one more step problems

Example 1 x2 + 8x = -

12 +12

+ 12

x2+ 8x + 12 = 0 ( x + 6) ( x + 2) = 0 ( x + 6 ) = 0 or ( x

+ 2) = 0 X = -6 or x = -2

Example 2 x2 + 2x = 63 -63 -

63

x2+ 2x - 63 = 0 ( x + 9) ( x - 7) = 0 ( x + 9 ) = 0 or ( x -

7) = 0 X = -9 or x = 7

At the end of your sheet of notes fill in the answers as we go, by setting the trinomials equal to zero, creating binomials and then setting each binomial = to zero.

Example 3 x2 - 14x + 60 =

12 -12

-12

X2 - 14x + 48 = 0 ( x - 8) ( x - 6) = 0 ( x - 8 ) = 0 or ( x -

6) = 0 X = 8 or x = 6