The Student Room Group
Reply 1
What can you tell me about quadratics and their roots?

Spoiler

0<x2m(x+1)=x2mxm=(xm2)2mm24 0 < x^2 - m(x+1) = x^2 - mx - m = (x - \frac{m}{2})^2 - m - \frac{m^2}{4} easier now? [critical values]
Reply 3
Do you mean that has to be positive?
Maybe consider the simon's hint or consider other values.
Reply 5
DeanK22
0<x2m(x+1)=x2mxm=(xm2)2mm24 0 < x^2 - m(x+1) = x^2 - mx - m = (x - \frac{m}{2})^2 - m - \frac{m^2}{4} easier now? [critical values]


could i not just look at the discriminant from there? i.e. where it is less than equal to zero
Reply 6
Just consider the critical values and use the table method
.
IBiot Ash '08
could i not just look at the discriminant from there? i.e. where it is less than equal to zero


CTS is just something I would do. You at liberty to use the discriminant if you please.
Reply 8
azhao
Just consider the critical values and use the table method
.


You lost me...I don't think I've done this yet, and even if i have it's after 10pm..I apologise for being particularly braindead....
Reply 9
IBiot Ash '08
You lost me...I don't think I've done this yet, and even if i have it's after 10pm..I apologise for being particularly braindead....


Use the discriminant. If it* is always positive, how many roots does it have?

*x^2-mx-m
SimonM
Use the discriminant. If it* is always positive, how many roots does it have?

*x^2-mx-m


2..But I can't see how that helps
Reply 11
IBiot Ash '08
2..But I can't see how that helps


Its always nonnegative

Edit: positive -> nonnegative

Latest