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Bunuel
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The way others have noted is "using Vieta's Equations in reverse," if you'd like to look into it. I try to avoid the memorization required.

Alternatively, we can assign values to constants and solve through.

A = 4

B = 16

C = -9

These can be any value, but I think making A anything besides +1/-1/0 makes it easier to understand.
Note* During a test, these can be adjusted if the original numbers you choose can't be easily factored.*

Therefore Equation one = 4x^2 + 16x - 9 = 0

Factor using the box method

(2x+9)(2x-1) = 0

Roots after solving = x = -9/2 , 1/2

the reciprocals of these are -2/9 and 2 which can be translated into: (9x+2)(-x+2) = 0

Foil that back out =.

-9x^2 + 18x - 2x +4
-9x^2 + 16x +4 =0

Replace assigned variables

4 = A

16 = B

-9 = C

Cx^2 + Bx + A = 0

Answer = A

Bunuel
What is the equation whose roots are reciprocals of the roots of \(ax^2 + bx + c = 0\)?

A. \(cx^2 + bx + a = 0\)

B. \(cx^2 - bx + a = 0\)

C. \(cx^2 + bx - a = 0\)

D. \(cx^2 - bx - a = 0\)

E. \(cx^2 + bx + a^2 = 0\)



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can anyone explain what the question means, I am not getting it
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The roots of an equation are its solutions (for example x1 and x2)

Their reciprocals are \(1/x1\) and \(1/x2\)

We're looking for another equation where the solutions are \(1/x1\) and \(1/x2\)

Hope that helps!
harjas2222
can anyone explain what the question means, I am not getting it
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