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secondary 3 | A Maths
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Nathan
Nathan

secondary 3 chevron_right A Maths chevron_right Singapore

Does anyone know part 3?

Date Posted: 1 month ago
Views: 17
J
J
1 month ago
Let the roots be α and β

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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Eric Nicholas K
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