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Find the sum of the squares of the roots of 3x^2 + 4x + 12 = 0.

 Mar 19, 2020
 #1
avatar+4569 
+1

We are asked to find: \(x^2+y^2\), for \(x\) and \(y\) are the roots of the equation.

\(3x^2+4x+12\) has a sum of \(\frac{-4}{3}.\)

\(3x^2+4x+12\) has a product of \(\frac{12}{3}=4.\)

 

Use the manipulation:\((x+y)^2-2(xy)=x^2+y^2\).

 

Thus, the answer is \((\frac{-4}{3})^2-2(4)=\frac{16}{9}-\frac{72}{9}=\frac{-56}{9}.\)

.
 Mar 19, 2020
 #2
avatar+111360 
+2

By Vieta....

 

We have  the form   Ax^2 + Bx + C  = 0

 

Let the roots  be  R1 and R2

 

The sum of these  =  -B/A  =   -4/3  =   R1 + R2     (1)

 

And the product of these =  C/A  =   =  12/3 =  4  = R1*R2  which implies that  8 = 2*R1*R2

 

Square  (1)  and we get that

 

R1^2  + 2*R1*R2  + R2^2  =  16/9

 

R1^2  + 8  + R2^2  =  16/9

 

R1^2  + 72/9  + R2^2  = 16/9      subtract 72/9  from both sides

 

R1^2 + R2^2  =  16/9 - 72/9  =  -56/9

 

 

cool cool cool

 Mar 19, 2020
 #3
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THX SO MUCH GUYS!!!!!!

 Mar 19, 2020

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