Can someone help me here?

[Neb]

Right, I'm doing some homework and I've got a couple of things that I'm not entirely sure if I've done correctly. Could you people show me how you'd do:

0 = 6x^2+6x-12

And

0 = 2x*9(1-x)^8

[Night Stalker]

Neb,

Just use addition/subtraction/mult/div to get x's on one side and and #s on the other. You are looking for all the solutions (where the eqn = 0) to each problem. Remember that what you do on one side, you need to do on the other. Factoring also helps.

I'll give a hint to start

6x^2+6x-12=0
---------- -
6......... 6

->

x^2+x-2=0

for this one there are 2 solutions

the other one there are 9

good luck

[Neb]

I've already done them, what I'm looking for is someone else to do them so I can see if I've done them correctly.

Forgot to add something. For the first the value of X must lie between -3 and 3. For the second the value of X must lie between 0 and 1. Also including the mentioned values, not just the ones lying in between.

[ 11-10-2002, 04:28 PM: Message edited by: Neb ]

[andrewas]

Originally posted by Neb: I've already done them, what I'm looking for is someone else to do them so I can see if I've done them correctly.

The general check is to put your values for X into the equation and evalutate. It should come out as zero. Im not solving a ^9 polynomial at this time of night.

[Vaskez]

Ah so you're Dracorion at Pandemonium? Didn't realise. I've answered this over at Pande but I'm so kind, I'll paste it here

6x^2 + 6x -12 = (3x - 3)(2x + 4) = 0
so the 2 roots are 3x = 3 so x = 1 and 2x = -4 so x = -2

You can easily check if you've done the factorisation right by multiplying out the brackets and you should get back to what you started. You can check that your x values are correct by putting them in 1 at a time and seeing if you get 0 as you expect. If you put x = 1 in then you get (3(1) - 3)(2x+4) =
0(2x+4) = 0 so you know x=1 is definitely a root. Do the same for the other value.

In the 2nd one you don't actually have to do any calculation, you can see by inspection that x = 0 and x = 1 are roots because substituting either of those into the equation gives you a value of 0.

[ 11-10-2002, 04:50 PM: Message edited by: Vaskez ]

[Vaskez]

Originally posted by andrewas: quote: Originally posted by Neb: I've already done them, what I'm looking for is someone else to do them so I can see if I've done them correctly.

The general check is to put your values for X into the equation and evalutate. It should come out as zero. Im not solving a ^9 polynomial at this time of night.[/QUOTE]You don't need to solve it, you can see by inspection that 2 of of the roots are 0 and 1 and the other roots should be repeat roots of x = 1 because it's (x-1)^8 if I remember by GCSE maths correctly!

it's quite easy actually [img]smile.gif[/img]

that formula can break down to:

6x(x+1) - 12 = 0
so
6x (x+1) = 12
or
x(x+1) = 2

therefore
x^2 + x = 2

so i think x=1 or x=-2

[Vaskez]

Originally posted by Lord Shield: it's quite easy actually [img]smile.gif[/img] that formula can break down to: 6x(x+1) - 12 = 0 so 6x (x+1) = 12 or x(x+1) = 2 therefore x^2 + x = 2 so i think x=1 or x=-2

why did you write all that out? I already did that ages ago [img]smile.gif[/img] look above

[LordKathen]

It' been way to long... I'm confused...

[LennonCook]

Originally posted by LordKathen: It' been way to long... I'm confused...

It`s easy- 6 steps:
Expand, Simplify, Factorise, Split, Solve, Check. [img]smile.gif[/img]