Solving Quadratics

By: Amanda & Melina

How Would You Solve the Equation: 3q² + 2q = 2q² - 2q + 45 (Difficult)

1. The first step of solving the equation is to get everything away from the right side.

Ex. (3q² + 2q = 2q² - 2q + 45)


a) Subtract 2q² from 3q² and getting q²

_3q² + 2q = 2q² - 2q + 45

-2q²

q² + 2q = 2q + 45


b) Next, add 2q to 2q , getting 4q

q² + 2q = -2q + 45

___ - 2q

q² + 4q = 45


c) Now, subtract 45, getting -45 at the END of the equation.

q² + 4q = 45

____-45

q² + 4q - 45 = 0


d) Your factored equation should be q² + 4q - 45 = 0


2. How to find the solutions of Quadratics

Ex. (q² + 4q - 45 = 0)


a) First step is to find 2 factors of -45 that have a sum of 4.


b) The 2 factors would be (9,-5)


c) Next, make 2 binomials

Ex. (q+9) (q-5) = 0


d) Once you have your 2 binomials, make both equations equal 0.

Ex. q+9=0 q-5=0


e) Solve the equations-

q + 9 = 0 (Subtract the 9 from both sides)

___- 9 q = - 9


q - 5 = 0 (Add 5 to both sides)

__+5 q = 5


f) Now you have found your solutions. (-9,5)

_______________________________________________________________________________________

**An easier way to find out what your solutions are...

1) When you find out the factors,

How would you solve the equation: k² + 13k + 40 = 0 (medium)

a) The first step multiply k² and 40 which gets you 40

+ 13k + 40 = 0


b) The second step is to find the factors of 40 that equal 13 which is 8 and 5

__40|13

8+5 | 13


c) So then you take the 13 and split it into two parts (using the two factors):

k² + 13k + 40 = 0

k² + 8k + 5k + 40 = 0


d) Now here is the tricky part... split the equations into two parts:

k² + 8k | 5k + 40 = 0

e) Then you find the common factor in both equations as so:

k² + 8k

*Think about it what do they both have in common. They both have k and 1 as factored the same so...*


k(k + 8)

*Use distributive property on the first equation in order to check if your right*

k(k + 8) = k² + 8k = 0


f) Do the same thing to the other half of the equation:

5k + 40 = 0

*Find what they both have in common. They are both divisible by five.*


5(k + 8)

*Always makes sure to check*

5(k + 8) = 5k + 40 = 0


g) Then plug the equations back together again:

k(k + 8) + 5(k + 8)

*Take the (k + 8) and add them with the (k + 5)


h) Solve both (k + 8) and (k + 5)

k + 8 = 0

___-8__-8

k = -8


i) Solve

k + 5 = 0

__- 5__-5

k = -5


You are left with your solution set which is {-5, -8}

Big image

How would solve the equation: (x - 2) (x + 6) = 0 (easy)

1. The first step is to set both of the equations to 0

Ex. x-2=0 x+6=0


a) Now solve...

x - 2 = 0 (Add 2 to both sides)

__+2

x=-2

x + 6 = 0 (Subtract 6 from both sides)

__- 6

x=6

b) Then you get your solution set {-2 , 6}

How do we use Quadratic Equations and why?

We use Quadratic Equations in the real world by...


Certain jobs use quadratic equations...


  • Astronomers- They use quadratic equations to describe the orbits of our planets, solar systems and galaxies. Physicists use quadratic equations to describe different types of motion. Even chemists need to use these equations in order to describe certain types of chemical reactions.