Yeah they do a bit...Perhaps I am going blind, but (d) and (e) look pretty similar to me...
Yeah they do a bit...Perhaps I am going blind, but (d) and (e) look pretty similar to me...
Yeah your right, I missed the 1 on the end, I will correct it.In part 3, should it the 2 equations be 2-2a-b=0 and 6-4a+b=0 , therefore a=2 and b=2?
You have to show the rest of the question to us - it could be anything.can u do (iii) plzzzz
Like AAEldar said, we need the rest of the question. Probably the easiest way to do it however (I'm guessing as I don't know the specific question), would be to manipulate the sums of roots inside each of the brackets. For example, if alpha+beta+gamma=9, then alpha+beta=9-gamma and then you can sub that in.can u do (iii) plzzzz
LOL, portal.However, this thing is most certainly a lie.
Thnxx broo<a href="http://www.codecogs.com/eqnedit.php?latex=\alpha @plus;\beta @plus;\gamma =0 \\ \therefore \alpha @plus; \beta=-\gamma \\ \alpha @plus; \gamma=-\beta \\ \gamma@plus; \beta=-\alpha \\ (\alpha@plus;\beta)^2(\gamma@plus;\beta)^2(\alpha@plus;\gamma)^2=(-\gamma)^2(-\beta)^2(-\alpha)^2 \\ =(-\alpha\beta\gamma)^2 \\ =(5)^2\\ =25" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\alpha +\beta +\gamma =0 \\ \therefore \alpha + \beta=-\gamma \\ \alpha + \gamma=-\beta \\ \gamma+ \beta=-\alpha \\ (\alpha+\beta)^2(\gamma+\beta)^2(\alpha+\gamma)^2=(-\gamma)^2(-\beta)^2(-\alpha)^2 \\ =(-\alpha\beta\gamma)^2 \\ =(5)^2\\ =25" title="\alpha +\beta +\gamma =0 \\ \therefore \alpha + \beta=-\gamma \\ \alpha + \gamma=-\beta \\ \gamma+ \beta=-\alpha \\ (\alpha+\beta)^2(\gamma+\beta)^2(\alpha+\gamma)^2=(-\gamma)^2(-\beta)^2(-\alpha)^2 \\ =(-\alpha\beta\gamma)^2 \\ =(5)^2\\ =25" /></a>
Sum of roots is -b/a. As the coefficient of x^2 is 0, the sum of roots is zero.Thnxx broo
How does alpha + beta + gamma = 0
:S
thnx mannn!Sum of roots is -b/a. As the coefficient of x^2 is 0, the sum of roots is zero.
parti (i) sonThnxx broo
How does alpha + beta + gamma = 0
:S