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secondary 3 | A Maths
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Does anyone know part 3?
Sum of roots = α + β = -b/a
Product of roots = αβ = c/a
Now we want roots that are the reciprocal.
So your roots are now 1/α and 1/β
Sum of roots = 1/α + 1/β
= (α+β)/αβ
= (-b/a)/(c/a)
= -b/c
Product of roots = 1/α × 1/β
= 1/αβ
= 1/(c/a)
a/c
So equation is :
x² +(b/c)x + a/c = 0
cx² + bx + a = 0
The a and c just switch positions, like you see in your answer in part ii).
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