Film-Tech Forum ARCHIVE: Attention Evans: Help with algebra!!!

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Posted by Andrew McCrea (Member # 674) on 09-20-2007, 01:30 PM:

Hello everyone,

Since my technical questions about computers have always been answered very well, I'm hoping that I could also get some help with some math equations. I have an exam on Monday worth 15% of my final mark in Business Mathematics.

$-$

I have no idea where to start.

Posted by Mike Blakesley (Member # 26) on 09-20-2007, 01:56 PM:

Business Math? That's kind of funny, I've been in business for over 30 years and I've never had to solve anything like that.

Also I sucked at algebra. I'll bet Evans Criswell will jump on this one... You should put "Attention Evans" in your title. [Big Grin]

Posted by Andrew McCrea (Member # 674) on 09-20-2007, 02:48 PM:

There, I edited it. I'm hoping for someone to walk me through it step-by-step.

I agree with you on the whole "suck at algebra/never have to use that" thing.

Ah, well.

Posted by William T. Parr (Member # 677) on 09-20-2007, 03:01 PM:

Use this for starters P lease E xcuse M y D ear A unt S ally. As In do anything in Parenthesis and Exponential wise then work out the Multiplication and Division and finish with the Addition and Subtraction part of the equation

Posted by Carl Martin (Member # 1146) on 09-20-2007, 03:04 PM:

what are we supposed to do with that expression? it isn't an equation...

Posted by Andrew McCrea (Member # 674) on 09-20-2007, 03:33 PM:

I think its simplification to find an answer. Like I said... Math has always been my weakest subject (and we never covered this in consumer).

Posted by Chris Slycord (Member # 4239) on 09-20-2007, 04:27 PM:

1)
1/2 * (3X^2 - X - 1) - 1/4 (5 - 2X - X^2)

2) Change the 2nd side to be the same order as you have in the first
1/2 (3X^2 - X - 1) - 1/4 (-X^2 - 2X + 5 )

3) Multiply the coefficients through
(3/2 X^2 - 1/2 X - 1/2) + (1/4 X^2 + 1/2X - 5/4)

3) Combine like terms
1.75 X^2 - 1.75

And I assume this is where they want you to find quadratic root which you solve for X. Then they want X-3 and you simply subtract and you've got your answer

Posted by Michael Cunningham (Member # 210) on 09-20-2007, 04:57 PM:

1.75x^2 - 1.75

Can you even find quadratic roots in this without knowing what the expression is equal to? Usually, an expression like this would have '= 0' tacked on to the end. Or are they asking you to sub in x - 3 for x and simplify the expression? Although that would be written more like 'For x = x - 3'.

Doing the quadratic formula for grins we get :

[-b + or - (b^2 - 4ac)^.5] / 2a =
[-{0} + or - ({0}^2 - 4{1.75}{-1.75})^.5] / 2{1.75} =
[+ or - (12.25)^.5] / 3.5 =
[+ or - (-3.5)] / 3.5 =
-3.5 / 3.5 or 3.5 / 3.5 =
-1 or 1

Those are the quadratic roots, given the expression is equal to 0. Then x as -1 or 1 would yield -4 or -2, I guess...

Posted by Chris Slycord (Member # 4239) on 09-21-2007, 12:52 AM:

quote: Michael Cunningham
Can you even find quadratic roots in this without knowing what the expression is equal to?

Technically, no but in the vast majority of cases you are given an equation and you find the nodes/zeros of the formula but the "=0" is implied. This is because you're trying to look for the cases where the formula is equal to zero.

Plus, I was going under the assumption that the formula he posted was missing the part where it says "Solve the following" then had that one posted where it's saying "solve [some equation] for X-3"

quote: Michael Cunningham
Or are they asking you to sub in x - 3 for x and simplify the expression? Although that would be written more like 'For x = x - 3'.

If it were a substitution case, it wouldn't say "for x-3" unless there were another formula he didn't post that had a variable named X then he is substituting the new formula he posted for x-3 into the other one.

Posted by Joshua Waaland (Member # 287) on 09-30-2007, 11:20 PM:

I got the same answer as Michael. Here is how I did it.

1) Set the quadratic expression equal to zero and make it a quadratic equation.

1/2(3x^2-x-1)-1/4(5-2x-x^2)=0

2) Multiply through the equation by 4 so you can use whole numbers and get rid of those ugly fractions. I learned this after a few semesters of math to make things easier.

2(3x^2-x-1)-(5-2x-x^2)=0

3) Distribute the two in the first part of the equation and the one in the second part of the equation remembering that it is actually a negative one.

6x^2-2x-2 (first part)
-5+2x+x^2 (second part)

gives you 6x^2-2x-2-5+2x+x^2=0

4) Combine like terms and you get...

7x^2-7=0

5) Divide both sides by seven and you get....

x^2-1=0

6) This is a difference of squares which is equal to..

(x+1)(X-1)=0

7) Solving for each possible x we get..

x=-1 and x=1

8) My guess is that the textbook is looking for just a plug the numbers in kind of thing for x. Then x-3 would give you...

-1-3 equals -4

1-3 equals -2

I took my algebra classes a few years ago and didn't use it much after we got into calculus stuff so I am a little rusty. Also the question is vague without knowing what you have been doing in class and the context of the situation.

I was on vacation this past week so I hope this helps you understand it even though your test is over.

BTW: For future reference I found the algebra website www.purplemath.com to be very helpful when I was taking algebra. It was written by a math teacher and has great info in it especially on those eleventh hour nights right before a test.