[Mr Nonsense]

revising for my further pure test at school on monday, and in the textbook there was this question:

consider the quartic equation

find a relationship between b and c which will ensure that the equation has one real root

I am not really sure at all what to do.... I have differentiated and found the one turning point (i don't think that there are more) -

and then this gives a horrible value for y, involving lots of cube roots, etc. but i don't really see where it is going. I understand where you have two or three turning points and you want one on either side of the x-axis to get 2 or 4 solutions, but here i have no idea.

So can anyone help? thanks in advance!

[SimonM]

One turning point, its going to be a minimum. Be careful when you substitute back in and it comes our reasonably neatly

[Swayum]

If it has one turning point and you want one real root, and given that a quartic starts high and ends high (like a parabola), where would be a cunning place to place the turning point in terms of the graph?

[somedumbname]

Edit: Never mind.

[generalebriety]

Originally Posted by somedumbname

If you want the full solution:

*snip*

Hope that helps.

Since you're new here, I should probably let you know that posting full solutions is typically frowned upon in here - see some of the sticky threads at the top of the maths forum for more information.

[somedumbname]

Whoops! Sorry about that

[Mr Nonsense]

Originally Posted by SimonM

One turning point, its going to be a minimum. Be careful when you substitute back in and it comes our reasonably neatly

Originally Posted by Swayum

If it has one turning point and you want one real root, and given that a quartic starts high and ends high (like a parabola), where would be a cunning place to place the turning point in terms of the graph?

Thank you both for your assistance - it is a great help!

I have substituted in the value for x at the turning point and y must = 0 for it to "sit" on the x-axis and this is the coniditon i got:

Originally Posted by somedumbname

Whoops! Sorry about that

is that the same as you were thinking?

can it be simplified much more?

thanks again! you are all so helpful

[Swayum]

No, that can't be right.

Choose b = 4, so

256c^3 = 63 * 256
c = cbrt(63) (that's the cube root of 63)

So y = x^4 - 4x + cbrt(63)

Turning point x = 1

But x = 1 clearly doesn't mean y = 0.

*Edit*

I make the answer as

Spoiler:

27b^4 = 256c^3 or if you prefer