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Let r and s be the roots of 3x^2 + 4x + 12 = 2x^2 + 3x + 11. Find r^2 + s^2.

Guest Feb 5, 2021

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Best Answer

+2401

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3x^2 + 4x + 12 = 2x^2 + 3x + 11

x^2 + x + 1 = 0

We're going to use vieta's formula. (https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas)

r^2 + s^2 = (r+s)^2 - 2rs

r^2 + s^2 = (-1)^2 - 2(1)

r^2 + s^2 = 1 - 2

-1

I hope this helped. :))

=^._.^=

catmg Feb 5, 2021

1⁺⁰ Answers

+2401

0

Best Answer

3x^2 + 4x + 12 = 2x^2 + 3x + 11

x^2 + x + 1 = 0

We're going to use vieta's formula. (https://artofproblemsolving.com/wiki/index.php/Vieta%27s_Formulas)

r^2 + s^2 = (r+s)^2 - 2rs

r^2 + s^2 = (-1)^2 - 2(1)

r^2 + s^2 = 1 - 2

-1

I hope this helped. :))

=^._.^=

catmg Feb 5, 2021

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