Ordinary Level Maths: Help Needed!

Keano!

Two quick questions I just can't do.

* = Squared
^ = Cubed



1st:

Solve for x and y

y=10-2x
x*+y*=25

I worked it down to be x=5 and y=0.

Hence, find the two possible values of x^+y^

Is it '-5^+0^=-125' and '5^+0^=125'
So the two values are -125 & 125?


2nd:

Let f(x)=x*+ax+t where a, t (are an element of) R

i) Find the value of a, given that f(-5) = f(-1)

I worked that down to be a=-6

ii) Given that there is only one value of x for which f(x)=0, find the value of t.

Can't get this at all



Please help, need this done for tomorrow morning!

Keano!

Bump?? Anyone.... really need this tonight

thetonynator

I've tried .. but i'm really confused . .. tiredness i think!

Keano!

Yeah, I'm confused as hell too
Thanks for trying though...


Anyone else able to help me?

rainbowtrout

The easy bit first, x = 3 and y = 4

y=10-2x
x*+y*=25

substitute 10-2x into the second equation for y

so X* + (10-2x)* = 25

multiply it out to get rid of the brackets:


x* + 4x* - 40x + 100 = 25

5x* - 40x +100-25= 0

5x* - 40x + 75 = 0

divide by 5 to simplify:

x* - 8x + 15 = 0

get factors for equation :

(x-3)(x-5)= 0

x = 3, x = 5

MathsManiac

Two wee errors there.

The squaring out should read 40x (corrected in the next line).

More significantly, the final factors should be (x-3)(x-5), giving x=3 or x=5.

For each of these values of x, there's a corresponding value of y, which you find using the first equation y=10-2x.

You had already y=0, corresponding to x=5.
The second solution, x=3, gives y=4.

The two solutions are: {x=5 and y=0} and {x=3 and y=4}

Each of these pairs can then be used to get a value for x^+y^.

The answers are 125+0=125 and 27+64=91.

MathsManiac

2nd problem:

So, after what you've done, you have f(x) = x* - 6x + t.

You're told f(x) = 0 has only one solution. There's a couple of ways to approach this, but here's one:

When you factorise a quadratic to solve it, you get (x - ?)(x - ?) = 0.
If there's only one root, it means that the same number has to be in both brackets.

Can you figure it out from there?

rainbowtrout

MathsManiac wrote: »

Two wee errors there.

The squaring out should read 40x (corrected in the next line).

More significantly, the final factors should be (x-3)(x-5), giving x=3 or x=5.

For each of these values of x, there's a corresponding value of y, which you find using the first equation y=10-2x.

You had already y=0, corresponding to x=5.
The second solution, x=3, gives y=4.

The two solutions are: {x=5 and y=0} and {x=3 and y=4}

Each of these pairs can then be used to get a value for x^+y^.

The answers are 125+0=125 and 27+64=91.

Hate typing maths. Always end up with a typo. :mad:

Keano!

MathsManiac wrote: »

2nd problem:

So, after what you've done, you have f(x) = x* - 6x + t.

You're told f(x) = 0 has only one solution. There's a couple of ways to approach this, but here's one:

When you factorise a quadratic to solve it, you get (x - ?)(x - ?) = 0.
If there's only one root, it means that the same number has to be in both brackets.

Can you figure it out from there?

-3?

making t=9?

Keano!

Oh, and many thanks for the solution to the first one

MathsManiac

Keano! wrote: »

-3?

making t=9?

Correct.

Keano! wrote: »

Oh, and many thanks for the solution to the first one

You're welcome!