-
We need your support!
We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
Search results
-
-
-
Further Mathematics: Post your doubts here!
Thanks for your help ;)- Thought blocker
- Post #683
- Forum: A-Levels
-
-
Further Mathematics: Post your doubts here!
As α, β and γ are the roots of the equation, (x - α)(x - β)(x - γ) = 0 Expanding the above, x³ - x²(α + β + γ) + x(αβ + βγ + γα) - αβγ = 0 ------- (1) But given is: x³ + x - 1 = 0 --------- (2) Comparing the corresponding coefficients of (1) & (2): (α + β + γ) = 0; (αβ + βγ + γα) = 1 and...- Thought blocker
- Post #681
- Forum: A-Levels
-
Further Mathematics: Post your doubts here!
1 / r(r + 1)(r - 1) = A/r + B/(r + 1) + C/(r - 1) such that the coefficients of terms in r² and r add to zero and the constant comes to 1. Then the numerator becomes A (r + 1) (r - 1) + B r (r - 1) + C r (r + 1) = Ar² - A + Br² - Br + Cr² + Cr. Thus, A + B + C = 0, C - B = 0, and -A = 1...- Thought blocker
- Post #680
- Forum: A-Levels
-
-
-
-
-
-
-
Finding someone! PK100 Candidate no: 280
I am the one you are finding. :D- Thought blocker
- Post #8
- Forum: Help, Ideas & Suggestions
-
-
-
-
-
-
-