If the sum of the roots of a quadratic equation is 3 and the sum of their cubes is 63 find the equation.

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Let x1 and x2

==> x1+x2 = 3.......(1)

==> x1^3+ x2^3 = 63.......(2)

from (2)

==> x1^3 + x2^3 = (x1+x2)^3 - 3ab(a+b)

==> 63 = 3^3 - 3ab(3)

==> 63 = 27 - 9ab

==> 36 = -9ab

==> ab = 36/-9 = -4

Now we know that the sum =...

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Let x1 and x2

==> x1+x2 = 3.......(1)

==> x1^3+ x2^3 = 63.......(2)

from (2)

==> x1^3 + x2^3 = (x1+x2)^3 - 3ab(a+b)

==> 63 = 3^3 - 3ab(3)

==> 63 = 27 - 9ab

==> 36 = -9ab

==> ab = 36/-9 = -4

Now we know that the sum = 3

and the product = -6

==> The equation is:

x^2 - 3x - 4 = 0

 

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