Math help needed (PERT pretest)

[Planr]

Hey guys I am taking the PERT test tomorrow morning for my community college stuff, and I am taking the free pretest handout packet they gave me at home. I need some help with a few of these math problems here:

x^2 - 6x + 5 = 0

I simplified this first to

x^2 - 6x = -5

and then finally to

x^2 = 6x -5

but now im stuck and have no idea what to do. I cant divide by X because there is no X on the -5. Im just stuck.


Im also having a problem with this other problem here.

Factor Competely:

x^2 - x - 6

The issue I am having with this one is that I have totally forgotten how to factor. I can easily do it the other way with FOIL or whatever that process is, but I don't remember how to change it into two separate sets of brackets. Some pictures of people doing this process on paper would help greatly.

Seventh Sandwich pls come to my rescue

[Ad Bot]

[Akio]

aren't you supposed to factor the x^2

and then you plug in to see which one works (can be both)

I have totally forgotten how to factor. I can easily do it the other way with FOIL or whatever that process is, but I don't remember how to change it into two separate sets of brackets. Some pictures of people doing this process on paper would help greatly.

Once you have the equation into:

x^2 - 6x + 5 = 0 you take the factors of 5 (which are 5, 1) and put one of each into a parenthesis like above

it works because -5 + -1 = -6, and -5 * -1 is +5
x^2 - x - 6 = 0
x(x - 6) = 0
x = 6

I don't know if that's correct, I haven't done this type of math in ages.

[ultimamax]

Yeah just factor
If you are bad at factoring then just use the quadratic formula

[Bloody Mary]

I like to use the quadratic formula for problems like that first one. As long as you have a calculator it's pretty easy to use.

[Katadeus]

It pays to get comfortable with factoring.

[Kyuande]

I was doing this recently, these links help me:
http://www.mathportal.org/calculators/solving-equations/quadratic-equation-solver.php
http://www.mathportal.org/calculators/polynomials-solvers/polynomial-factoring-calculator.php

These also go step to step to make sure you are getting it correctly.

[phflack]

x^2 - x - 6 = 0
x(x - 6) = 0
x = 6

I don't know if that's correct, I haven't done this type of math in ages.

x^2 - x - 6 = 0
(x + 2)(x - 3) = 0
x = -2, 3

x(x - 6) = x^2 - 6x

[Akio]

oh yeah i guess you can get -1 from 2 and 3

i was thinking of the procedure you do when it's something like

x^2 + 5x = 0
if the equation is something like

2x^2 + 18 - 36 = 0 (probably wont work, made up numbers)

Or if there's a greatest common factor, you can divide by that to make it MUCH easier.

So you would turn the above equation into:

x^2 + 9 - 18 = 0

[SWAT One]

[Ravencroft·]

What exactly is the pert test? This is like 9th grade algebra.

[Xalos]

2x^2 + 18 - 36 = 0

That's just x² - 9, or (x - 3)(x + 3).  x = ±3.  It's a difference of squares, one of the easiest possible quadratic forms to factor.

[keesfani]

Akio is right with the first one.

x^2 - x - 6 = 0
(x-3)(x+2) = 0
x=3 V x=-2

[Planr]

What exactly is the pert test? This is like 9th grade algebra.

Its the college readiness test you have to take when you go into college. Everyone has to take it, at least where I live. It helps the college people know how to place you within the college when it comes to courses.

[Klarck]

You should memorize those identities:

a^2 - b^2 = (a - b)(a + b)

(a + b)^2 = a^2 + 2ab + b^2

(a - b)^2 = a^2 - 2ab + b^2

[Katadeus]

also you might as well remember A^3 - B^3 = (A - B)(A^2 + AB + B^2) and A^3 + B^3 = (A + B)(A^2 - AB + B^2)
These are unlikely to come up as much though