We are asked to find: x2+y2, for x and y are the roots of the equation.
3x2+4x+12 has a sum of −43.
3x2+4x+12 has a product of 123=4.
Use the manipulation:(x+y)2−2(xy)=x2+y2.
Thus, the answer is (−43)2−2(4)=169−729=−569.
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