If α and β are roots of x\(^2\) + mx - n = 0, where
FURTHER MATHEMATICS
WAEC 2022
If α and β are roots of x\(^2\) + mx - n = 0, where m and n are constants, form the
| equation | whose | roots | are | 1 α | and | 1 β | . |
- A. mnx\(^2\) - n\(^2\) x - m = 0
- B. mx\(^2\) - nx + 1 = 0
- C. nx\(^2\) - mx + 1 = 0
- D. nx\(^2\) - mx - 1 = 0
Correct Answer: D. nx\(^2\) - mx - 1 = 0
Explanation
x\(^2\) + mx - n = 0
a = 1, b = m, c = -n
α + β = \(\frac{-b}{a}\) = \(\frac{-m}{1}\) = -m
αβ = \(\frac{c}{a}\) = \(\frac{-n}{1}\) = -n
the roots are = \(\frac{1}{α}\) and \(\frac{1}{β}\)
sum of the roots = \(\frac{1}{α}\) + \(\frac{1}{β}\)
\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{α+β}{αβ}\)
α + β = -m
αβ = -n
\(\frac{α+β}{αβ}\) = \(\frac{-m}{-n}\) → \(\frac{m}{n}\)
product of the roots = \(\frac{1}{α}\) * \(\frac{1}{β}\)
\(\frac{1}{α}\) + \(\frac{1}{β}\) = \(\frac{1}{αβ}\) → \(\frac{1}{-n}\)
x\(^2\) - (sum of roots)x + (product of roots)
x\(^2\) - ( m/n )x + ( 1/-n ) = 0
multiply through by n
nx\(^2\) - mx - 1 = 0
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