Recent content by Porty

ha ha.. Thank you! That works sweet... I wish i could think of doing stuff like that on my own... anyways here it is.. I took the sum: \alpha + 4 = k + 2 \alpha = k - 2 and put it in the product: \alpha * 4 = k - 4 and it gave me exactly what i needed... 4(k-2) = k - 4 I...

Oh ok.Thanks.. Um ok here's what i got the second equation from... The question is: For what value of k will the equation x^2 - (k+2)x + (k-4)=0 have: a) one root equal to zero. b) one root equal to 4? c) one root which is the reciprocal of the other? a) was easy c) was easy...

Good guess... Yep NSW Australia.. haha Yeah i knew that i did't have to figure out the roots themselves to solve it so i wasn't going to attempt it.. Your right, using the products of the roots did end up being enough to solve the product of the roots equation. I did'n even think of...

Thanks for the replies guys.. Just quickly leting you guys know why I am studying this.. I hope to get into uni next year, since i didn't do year 11 and 12 i decided to pick up a book on 2 unit math to prepare myself for what may be ahead... So some of my questions may seem really easy...

This is probably very easy, but anyway.. I just don't really understand the question, therefore don't know what answer to give.. Homework Statement The question reads... For what values of n will one root of the equation (n - 2)x^2 + (n + 2)x + 2n + 1 = 0 be the reciprocal of the...

HallsofIvy... Yep yep that's it... -x^2 + 4x + 5 - k = 0 = 4^2 - 4(-1)(5-k) = 36 - 4k 4k = 36 k = 9 Now i can sleep... Thank you

I know this is probably really easy, but i just can't figure it out.. Homework Statement Find values of k for which (5 + 4x - x^2 = k) has equal roots. I know the answer is k = 9 but i don't know how to get there.. Homework Equations ax^2 + bx + c = 0 Discriminant = b^2 - 4ac...