Higher Level Maths Thread

OceanSixteen

I'm wondering does anyone know how to solve this

(x^2 + ax - b) / (x^3 - 2bx^2 + ax - 6)

Show that: a = -2b^2 - 6/b

Hon the Dubs

I may be wrong but I'm nearly certain that the first fraction must be equal to something otherwise it can't be done

DublinArnie

I tried doing the question twice with two different methods, ain't working. Probably didn't cover this yet in class.

mayogirlie1

Do ya not just do the long division thing

mayogirlie1

Do ya not just do the long division thing

is mise spartacus

I tried long division but nope something isn't working for some reason

mr.turtle

OceanSixteen wrote: »

I'm wondering does anyone know how to solve this

(x^2 + ax - b) / (x^3 - 2bx^2 + ax - 6)

Show that: a = -2b^2 - 6/b

I think I have it
I divided x^2 +ax-b into the x^3 - 2bx^2 + ax - 6 and got x(a+b+2ab+a^2) and -6-2b^2 -ab at the bottom of the division if you get me . Then using -6-2b^2 -ab =0,
-6-2b^2= ab , then when you divide both sides by b you get -2b^2 -6/b. Hope that helped

is mise spartacus

mr.turtle wrote: »

I think I have it
I divided x^2 +ax-b into the x^3 - 2bx^2 + ax - 6 and got x(a+b+2ab+a^2) and -6-2b^2 -ab at the bottom of the division if you get me . Then using -6-2b^2 -ab =0,
-6-2b^2= ab , then when you divide both sides by b you get -2b^2 -6/b. Hope that helped

Now I know where I went wrong we've all learned something

Peregrine

mr.turtle wrote: »

I think I have it
I divided x^2 +ax-b into the x^3 - 2bx^2 + ax - 6 and got x(a+b+2ab+a^2) and -6-2b^2 -ab at the bottom of the division if you get me . Then using -6-2b^2 -ab =0,
-6-2b^2= ab , then when you divide both sides by b you get -2b^2 -6/b. Hope that helped

You've completely lost me..

Why are you diving the bottom of the fraction by the top? Why did you isolate a part of the remainder and why did you let it equal to zero?

mr.turtle

Nimr wrote: »

You've completely lost me..

Why are you diving the bottom of the fraction by the top? Why did you isolate a part of the remainder and why did you let it equal to zero?

I just took a chance dividing the top into the bottom fraction, I guess I must have done some question like it before . We were thought that each part of the remainder, like anything contains x or anything containing no variable was a seperate term and each of these was equal to zero

MathsManiac

The question as stated is completely meaningless. I STRONGLY suspect that it is NOT the way it was phrased in the book, exam paper, or notes that you found it in.

The fact that Mr. Turtle managed to get the answer suggests that what the question should have said is something like this:

"Given that (x^2 + ax - b) is a factor of (x^3 - 2bx^2 + ax - 6),
show that: a = -2b^2 - 6/b."

or perhaps:

"(x^2 + ax - b) divides (x^3 - 2bx^2 + ax - 6).
Show that: a = -2b^2 - 6/b ."

If the original question was one of these or similar, then what Mr Turtle did makes sense. If one expression divides evenly into another, then that means that when you do the division, the remainder is 0.

DublinArnie

MathsManiac wrote: »

The question as stated is completely meaningless. I STRONGLY suspect that it is NOT the way it was phrased in the book, exam paper, or notes that you found it in.

The fact that Mr. Turtle managed to get the answer suggests that what the question should have said is something like this:

"Given that (x^2 + ax - b) is a factor of (x^3 - 2bx^2 + ax - 6),
show that: a = -2b^2 - 6/b."

or perhaps:

"(x^2 + ax - b) divides (x^3 - 2bx^2 + ax - 6).
Show that: a = -2b^2 - 6/b ."

If the original question was one of these or similar, then what Mr Turtle did makes sense. If one expression divides evenly into another, then that means that when you do the division, the remainder is 0.

Yeah, you're right. OP didn't tell us the question in its proper form.

d1234

Are any of you guys nearly finished the maths course? What did ye find most challenging??

We have about 65% done. So far differentiation was hardest IMO...

mayogirlie1

Any one have any idea how to do this
A parabola has x axis intercepts of 6 and -3 and passes through point (1, 10) find equation of the parabola?

Ompala

mayogirlie1 wrote: »

Any one have any idea how to do this
A parabola has x axis intercepts of 6 and -3 and passes through point (1, 10) find equation of the parabola?

Parabola will have a characteristic equation y = ax^2 + bx + c, intercepts of the x axis imply y=0
We are told that at:
x = 6, y = 0
x=-3, y =0
x = 1 y =10

Substituting these into y = ax^2 + bx + c gives
36a + 6b + c = 0
9a -3b + c = 0
a + b + c = 10

Solving all 3 simultaneous equations gives a = -0.5, b =1.5, c = 9 so the answer is y = -0.5x^2 + 1.5x + 9

mayogirlie1

You genius thank you so much!

Ompala wrote: »

Parabola will have a characteristic equation y = ax^2 + bx + c, intercepts of the x axis imply y=0
We are told that at:
x = 6, y = 0
x=-3, y =0
x = 1 y =10

Substituting these into y = ax^2 + bx + c gives
36a + 6b + c = 0
9a -3b + c = 0
a + b + c = 10

Solving all 3 simultaneous equations gives a = -0.5, b =1.5, c = 9 so the answer is y = -0.5x^2 + 1.5x + 9

Take Your Pants Off

Sorry to spoil thread but any help in q27.

Moody_mona

Try dividing the quadratic into the cubic (it will help to write the cubic as x^3 + 0x^2 +0x + k) it will get messy but work to the end.

Because it divides in evenly (from the question) let your remainder equal zero.

Take Your Pants Off

Moody_mona wrote: »

Try dividing the quadratic into the cubic (it will help to write the cubic as x^3 + 0x^2 +0x + k) it will get messy but work to the end.

Because it divides in evenly (from the question) let your remainder equal zero.

Yeah wrote the cubic as that but is there no other way than dividing, as for the previous questions I used a different way and that worked, but I just cant seem to use it for this.
Thanks