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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.

 Aug 30, 2020
 #1
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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.

 Aug 30, 2020
 #2
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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

 Aug 30, 2020
 #3
avatar+8342 
+1

Consider \((r + s)^2 = r^2 + s^2 + 2rs\).

 

\(r + s = -\dfrac43\\ rs = 4\)

 

So

\(\left(-\dfrac43\right)^2 = (r^2 + s^2) + 2\cdot 4\\ r^2 + s^2 = -\dfrac{56}9\)

 Aug 31, 2020

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