Hi Guys

Im stuck on the following question

3x^2+16x-12=0 hence solve 3(y-(7/y))^(2)+16(y-(7/y))-12=0

I can get the factors (3x-2) and (x+6) for the first one and so x=2/3 x= -6

its the second part im stuck on

any help would be great please

Thanks
Pete

2. ## Substitue

In the second equation, let x=y-(7/y)
so u get the first equation by substitution of x
so
y-(7/y)=2/3
and y-(7/y)= -6
two more quadratics for you to solve!
(multiply throughout by y)

3. im still confused about the 2nd part

y-(7/y)=2/3
and y-(7/y)= -6
you have been given the answer here, have a go at solving these equations.

5. so the bottom one would be y^(2)-7y=-6y? and then y^(2)-7y+6y=0? and then y^(2)-y=0?

not sure about the 2/3 part though

6. Multiply y through both sides you get

$\displaystyle y^2-\frac{2y}{3}-7$

and

$\displaystyle y^2+6y-7$

7. I think the point is that y-(7/y) is the same as x in the first equation, therefore the results are the same - you have solved the first one, so by default you haver also solved the second

8. hmmm ok thanks for that

where did i go wrong on the one above (post 5)

9. so if y-(7/y) = x do i still have to solve that part of equation? or just go into 3(2/3)^(2)+16(-6)-12=0?

10. ## Clarity

Ok... now lets try to make things clear:
first of all there were two equations, first one which u already solved and found two values of x.

The second equation ie the one that contains y, requires a bit of common sense.

If you write both equations on paper in maths form, you will notice that the second one
just has y-(7/y) instead of x.

So... your first step will be to solve the second equation for y-(7/y).
Look, if you change x for z in the first equation, doesn't matter, does it?

So similarly u r just using y-(7/y) instead of x. But you already did that in the first equation.
And thats where you get the two new equations
y-(7/y)=2/3
and y-(7/y)= -6

But your objective of the second equation is NOT to find y-(7/y) which is x. It is to find y. It is these two new equations that u now need to solve, forget the original second equation, because these two equations are parts of the second one.

11. ## ah now i see

ahh thanks zeeshahmad now i kinda feel a little silly but at least i think i understand now.

your a star, ill post back what i get when i work those equations out.

Thanks again

Pete

12. so now i get y-(7/y) = -6, so y^(2) -7 = -6y, so y^(2)+6y-7=0, so (y+7)(y-1) =0 so y=-7 y=1

y-(7/y)= 2/3, y^(2)-7=2y/3 so multiply out by 3 gives 3y^(2)-21=2y, so 3y^(2)-2y-21=0 so (3y+7) (y-3)=0 so y = -(7/3) y=3?

Yup