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Let r and s be the roots of  \($3x^2 + 4x - 12 = 0. \)

 

Find \(r^2 + s^2\)

 Aug 6, 2022
 #1
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Sum of the roots = r + s  =    -4/3   square both sides

r^2 + 2rs + s^2  =  16/9

Product of the roots =  rs =  -12/3 = 4

2rs =  8

 

So

 

r^2 + 2rs + s^2 =  16/9

 

r^2 + 8  + s^2  =  16/9

 

r^2 + s^2 =   16/9   - 8

 

r^2 + s^2  = -56/9

 

 

cool cool cool

 Aug 6, 2022

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