SOLVE FoR X?

Question

x^2-5x=12

x^2-8x+15

16x^2
----------
24xy^3


25-n^2
-------------
n^2+n-30

Answer 1

1.) x^2 - 5x -12 = 0
Use quadratic formula as u can't factor this out.
x = {-b +/- (b^2 - 4ac)^1/2}/2a

x = {5 +/- (25 + 48)^1/2}/2
x = {5 +/- (73)^1/2}/2

2.) x^2 -8x + 15 = 0
x^- 5x - 3x + 15 = 0
(x - 5)(x - 3) = 0
x = 5, 3

Answer 2

Because you can't get an x alone on one side, you need to set the equation equal to 0 and factor.
x^2-5x-12=0 Then set up your parentheses
(x+_)(x-_) You know that one is positive and one is negative because the last term is negative
Now find two numbers that add to equal -5, and multiply to -12. There are no integers that do, so we need to use the formula (-b +/- sqrt(b^2-4ac))/2a.
a=1,b=-5,c=-12 so...
(5 +/- sqrt(25-4(1)(-12))/2(1) or (5 +/- sqrt(73))/2 and that is really all you can do with that.

Is the next one equal to 15 or 0? If it is 0:
Set up parentheses again:
(x-_)(x-_) This time they are both negative because they need to add to be negative and multiply to be positive
Then find two numbers who's product is 15.
1 and 15 , 3 and 5 , -1 and -15 or -3 and -5.
We already figured out that they are both negative, so it has to be -3 and -5.
(x-3)(x-5)

I am guessing for the second two you need to simplify.
16x^2/24xy^3 Because they are just simple monomials (nothing is being added) just simplify each part.
16/24=2/3 giving you 2x^2/3xy^3
x^2/x=x giving you 2x/3y^3 , and that is all you can do with that

The next one is a little harder. First you need to factor
25-n^2 is the difference of two squares, so you get (5-n)(5+n)
n^2+n-30 factors into (n-6)(n-5)
When you have (5-n)(5+n) / (n-6)(n-5) the (n-5)'s cancel out giving you (5+n)/(n-6)

Answer 3

x = 9

Answer 4

x^-5x-12=0

use qudratic formula a=1 b=-5 c=-12
x=(-b+/-(b^2-4ac)^0.5)/2a

plug in and solve using a calculator

and the second problem should be equal to zero to get:
(x-3)(x-5)=0 so x = 3 or x=5