DU mathematicians, help please...

[sakabatou]

I got three problems that, unofrtunately, I've forgotten how to solve. So, may I have some help?

The Problems (solve for x):
#1: x^2+14x+33=0
#2: 2x^2=5x-12=0
#3: 2x^2-9x+10=0

And no, I odn't want the answer, I want steps on how to solve.

[tjwmason]

1. There's a really simple formula to use

If one consider them to be of the form ax^2+bx+c=0

It goes something like

x=(-b+or-sqrt(b^2-4ac))/2a

But I'd check the final result as that formula is from memory, and I haven't had to use it since maths G.C.S.E. in 1995.

[sakabatou]

6. I cannot belive I forgot that!

x = (-b+or-sqrt((b^2-(4ac)))/2a

AUGH!

[Nye Bevan]

2. Here you go

#1

x^2 + 14x + 33 = 0
(x + 11)(x + 3) = 0
so x = -11 or -3


#2

2x^2 - 5x - 12 = 0
x^2 - 2.5x - 6 = 0
(x - 4)(x + 1.5) = 0
so x = 4 or -1.5

#3

2x^2 - 9x + 10 = 0
x^2 - 4.5x + 5 = 0
(x - 2.5)(x - 2) = 0
so x = 2.5 or 2

[Spinzonner]

5. Good math

Bad reading.

OP Didn't want the answers, wanted the method.

[sakabatou]

7. Don't give me answers!

:: hides eyes ::

[Nye Bevan]

14. Sorry--- here is the solution process

First step is to get the coefficient of x^2 to be 1.
This is not necessary for #1, as it is already 1, and is achieved for #2 or #3 by dividing throughout by 2.

Now we have an equation of the form x^2 + ax + b = 0.
We would like to factorize this as (x - m)(x - n) = 0, for some m and n, because then we know that either x-m = 0, so x=m, or x-n=0, so x=n.

So the question is how to do this factorization. Since

(x-m)(x-n) = x^2 - (m+n)x + mn

we need to find m and n so that mn=b and (m+n) = -a. These problems are simple enough that m and n can easily be found by trial and error.

A few respondents have mentioned the quadratic equation formula, which would certainly work but is overkill in this case (and almost certainly not what the questioner intended, due to the ease of factorization).

[Spinzonner]

3. Problem #2 seems to have an extra '=' sign

probably should be a '-'.

Quadratics will have two roots.

Either use the Quadratic Equation

http://mathworld.wolfram.com/QuadraticEquation.html

or factor the equations and extract roots from the factors.

[Robb]

4. All I remember is

negativebplusorminusthesquarerootofbsquaredminus4acallover2a.

Ask me if I remember what it's for. :D

[indy_azcat]

8. Wow... all I could think of was FOIL...

firstoutsideinsidelast :P

but at least I can kinda remember it's use :D

[tjwmason]

11. Isn't it pythagorous theorem

that we're all supposed to remember - without remembering what the hypoenteuse is, or its applicability?

[diddlysquat]

9. Try the quadratic formula:

aX^2 + bX + c = 0

Make sure the equation is set equal to zero before you start.
a is the number in front of X^2
b is the number in front of X
c is the number not in front of a variable

X = (-b plus or minus the square root of b^2 minus 4ac) divided by 2a

I don't have the symbols to actually write the equation.

Plug in a, b, and c and solve.

[Salviati]

10. So the trick to doing these is to factor them:

we want to factor them into the product of two things, call them a and b, because then, we'll have an equation that looks like:

a*b=0

which can only be true if either a=0 or b=0, so we can break the single quadratic equation into two linear ones, e.g.:

x^2+14x+33 = 0

This is the tricky part, essentially we want to find two numbers that will add to the coefficent on the x term, and multiply to be the constant term.

The fact that the constant term is postitive tells us that the two numbers we're looking for are both the same sign, as they need to multiply to a postive number.

The fact that the coefficent on x is positive, tells us that the two terms are going to be postive, because if two numbers are the same sign, and they add to a positive number, both must be positive

So, two numbers that will do the trick are 3 and 11, they add to 14 and multiply to 33. So factoring, we get the following product:

(x+11)(x+3)=0

so either

x+11 = 0 => x=-11

or

x+3 = 0 => x=-3


hopefully that helps, the other two are a bit trickier to factor, since they have a coefficent on the x^2 term, but that just means a bit more guessing and checking during the factoring part, the stuff about the signs still works in that case...

[Floogeldy]

12. You people intimidate the hell out of me.

;)

[sakabatou]

13. Thank you all! I got 'em!

YOu don't have to reply anymore! In other words... Locketh this topic, please.

[Orsino]

15. Five easy steps.

1. Keep repeating that the equations have turned the corner.
2. Blame x's pessimism for the equations' not solving themselves.
3. Fiddle about with unrelated equations.
4. Bomb b, and blame the attacks on c.
5. Return to step 1.

[bicentennial_baby]

16. ....

:rofl: i wish i could use that in my next algebra tutoring session...

[miss_kitty]

17. What the hell

does ^ mean?
Sincerely,
MathStupe:dunce:

[steely]

18. ^ means to raise to the power of....

so x^2 means x raised to the 2nd power, or x squared

you can demo in excel

sometimes you may see x**2

[miss_kitty]

19. Oh

So it's because you can't write a tiny 2 next to the x on these magic boxes we all use. Well, colour me astounded.:wow:

[steely]

20. you're right - older computers wouldn't support superscripts

I misunderstood where you were coming from.

[miss_kitty]

21. no I did not know what it was

I Am the Stupe of math and I have the grades to prove it. I thought it was funny that with all the things these computers can do, describing exponential factors the way we were taught in school isn't possible!

:D