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## Question

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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