You can put this solution on YOUR website! The sum of the roots of equation is 2 and sum of their cubes is 98 than find the equation
let roots are A and B
sum of roots A +B =2 (1)
A^3 +B^3 =98 we know that A^3 +B^3 =(A+B)(A^2+AB +B^2) by substitution
(A+B)(A^2-AB +B^2)= 98
2(A^2-AB +B^2)=98
(A^2-AB +B^2)=98/2 =49 we know that A^2 + B^2 = (A+B)^2-2AB
A^2 + B^2 +AB =49
(A+B)^2-2AB -AB =49
2^2-3AB=49
3AB = -45
AB=-15
equation formation is X^2-(sum of roots )X +product of roots
X^2-2X +(-15)=0
x^2-2x-15=0
