Solve the equation `x^3-13 x^2+15 x+189=0` if one root exceeds the other by 2.

Explanation

Let the roots be `alpha, alpha + 2, beta`. Sum of roots is `2alpha + beta + 2 = 13`.
`therefore beta = 11 - 2alpha ` (1)
Sum of the product of roots taken two at a time is
`alpha (alpha + 2 ) + (alpha + 2) beta + beta alpha = 15`
or `alpha^(2) + 2alpha + 2(alpha + 1) beta = 15` (2)
Product of the roots is
`alpha beta (alpha + 2) = - 189` (3)
Eliminating `beta` from (1) and (2) , we get
`alpha ^(2) + 2alpha + 2(alpha + 1) (11 - 2alpha) =15`
or `3alpha^(2) - 20 alpha -7 =0`
or ` (alpha - 7 ) (3alpha + 1) = 0`
or `alpha = 7 or -(1)/(3)`
`rArr beta = - 3, (35)/(3)`
Out of these values , `alpha = 7, beta = - 3` satisfy the third relation `alpha beta (alpha + 2) = - 189, i.e., (-21)(9) = -189`. Hence, the roots are
`7, 7 + 2, -3 or 7, 9, -3.`