functions of equations

[mollykins]

if you only know a and b and what f(x) equals, how do you work out what x is?

[Count-A-Strong]

F U N E M?

[boxtops]

Give us an example - a hypothetical question is too much for me at this time of day.

[mollykins]

a= -3 (I think) and b=7 (I think) and f(x) = 35

starting with f(x)= x^2 +ax +b I get up to 28 = x^2 + -3(x)

[mollykins]

because there isn't a c, you can't use the quadratic formula can you to get x, there's too many unknowns isn't there?

[jake-the-peg]

you have a function f(x)=x²-3x+7 and f(x)=35

so 35=x²+3x+7

x²+3x-28=0

simple quadratic

[mollykins]

would you have to do, 35^2+35(-2)+7 ???????? That's jsut a guess, but the numbers seems really wrong plus I can't get a positive and negative number from that

[mollykins]

jake but you have to find the positive and negative value of x

[mollykins]

and wouldn't it be x^2-3x-28=0???

[Fair-Stumped]

Mollykins, your last statement is correct. You should be able to solve it and come up with an integer as the value of x. There is an integer value for x which solves the equation but I'm not going to tell you what it is.

[Fair-Stumped]

Of course there is also a negative value which solves the equation

[factor-fiction]

You can simply factorise it mollykins.
If it is x²-3x-28=0 you need two numbers which multiply together to give -28 and add together to give -3. That is -7 and +4.
(x-7)(x+4)=0
So that leads to the values of x

[Prudie]

In my day a was the coefficient of x^2 and b the coefficient of the x term I'm sure it hasn't changed?
eg -3x^2 + 7x - 35 = 0

[vascop]

Mollikins wrote:
because there isn't a c, you can't use the quadratic formula can you to get x, there's too many unknowns isn't there?

Yes you can, but its so simple to solve when c=0, because ax^2+bx=0,
so x(x+b)=0
so x=0 and x=-b/a are the two roots in this case
but the normal formula works as well:
x=(-b+/-sqrt(b^2-4ac))/(2a)
When c=0 this becomes
x=(-b+/-sqrt(b^2))/(2a)=(-b+/-b)/(2a)
so x=(-b-b)/(2a)=-2b/2a=-b/a
and x=(-b+b)/(2a)=0
Same as above.

[mollykins]

aaahhhhhh of course, thanks.

[factor-fiction]

Vascop is correct although when he/she said
"Yes you can, but its so simple to solve when c=0, because ax^2+bx=0,
so x(x+b)=0"
I think that should have read
so x(ax+b)=0

You are doing well to stick with Maths, mollykins.

Most secondary pupils are totally bamboozled by algebra and just cannot follow it, however teachers try to approach it.

[mollykins]

My mum thinks i'm wierd for choosing to do maths . . . .

[vascop]

Thanks for the correction factor. I'm a he by the way.

[vascop]

Molly
I can understand why you chose maths. It's a fun subject, as well as being challenging and useful. What's wrong with being weird anyway? Better than being boring.