How to deal with tricky quadratic and cubic equation
So today lets discuss how to solve the problems related to quadratic and cubic equations. Though these problems are simple but a bit tricky. Similar approach is required in each and every problem.
First lets discuss about “sum of roots” and “product of roots” of a polynomial.
Considering a polynomial ax^n + bx^n-1 + cx^n-2 ……. +z =0
Sum of roots = -b/a
Sum of roots taking two at a time = (-1)^n-1 (c/a)
Product of roots = (-1)^n (z/a)
Now Considering quadratic equation
ax^2 + bx + c = 0
Sum of roots = (x1 + x2) = -b/a
product of roots = (x1.x2)= c/a
Considering cubic equation
ax^3 + bx^2 + cx + d = 0
Sum of roots = (x1 + x2 + x3) = -b/a
sum of roots taking two at a time = (x1.x2 + x2.x3 + x3.x1) = c/a
product of roots = x1.x2.x3 = -d/a
Example : Each root of the equation ax3 – 7×2 + cx + 231 = 0 is an integer. One of the roots is = -1/2 times the sum of the other two roots. What is the sum of all the possible values of a?
Solution : Let the roots be x1, x2, x3.
so x1 = 1/2(x2+x3)
Now considering given cubic equation.
x1 + x2 + x3 = – (-7/a) = 7/a —————–i
x1.x2.x3 = -231/a ————————-ii
Also x1 = ( -1/2).(x2 + x3) —> (x2 + x3) = -2×1
So from (i)..
x1 – 2×1 = 7/a -> x1 = -7/a
also x1.x2.x3 = 231 /a — > (-7/a )x2.x3 = 231/a — > x2.x3 =33
So we can have following values of x1 , x2 and x3
x2(x3) x3(x2) x1
33 1 -1/2(33+1) = -17
11 3 -1/2(11+3) = -7
-33 -1 -1/2(-33-1) = 17
-11 -3 -1/2(-11-3) = 7
so the possible values of a = -7/x1 = 7/17 or 1 or -7/17 or -1
Hence the sum of all possible values of a = 7/17 + 1 -7/17 -1 = 0.