how do you find r^2 + s^2 if you only have r+s and rs?

(problem I'm trying to solve:)

Let r and s be the roots of 3x^2 + 4x + 12 = 0. Find r^2 + s^2.

Guest Aug 30, 2020

#1**+1 **

The roots of the quadratic are -(2^(5/2)*i + 2)/3 and (2^(5/2)*i - 2)/3, so r^2 + s^2 = (-(2^(5/2)*i + 2)/3)^2 + ((2^(5/2)*ii - 2)/3)^2 = 23/9.

Guest Aug 30, 2020

#2**0 **

Solve for x:

3 x^2 + 4 x + 12 = 0

Divide both sides by 3:

x^2 + (4 x)/3 + 4 = 0

Subtract 4 from both sides:

x^2 + (4 x)/3 = -4

Add 4/9 to both sides:

x^2 + (4 x)/3 + 4/9 = -32/9

Write the left hand side as a square:

(x + 2/3)^2 = -32/9

Take the square root of both sides:

x + 2/3 = (4 i sqrt(2))/3 or x + 2/3 = -(4 i sqrt(2))/3

Subtract 2/3 from both sides:

x = (4 i sqrt(2))/3 - 2/3 or x + 2/3 = -(4 i sqrt(2))/3

Subtract 2/3 from both sides:

**r = (4 i sqrt(2))/3 - 2/3 or s= -(4 i sqrt(2))/3 - 2/3**

**(2/3 (-1 - 2 i sqrt(2)))^2 + (2/3 (-1 + 2 i sqrt(2)))^2 = - 56 / 9**

Guest Aug 30, 2020