Sum of Roots Formula

Let us consider a quadratic equation ax2 + bx  + c = 0
Since the equation has the power 2, it has two roots.  Let them be α and β. Then sum of the roots is  α + β.
The roots of the quadratic equation is given by the formula x = - b ± √ b2 - 4ac
2a
Then α = -b + √ b2 - 4ac     and   β =  -b -  √ b2 - 4ac
2a   2a
Then α  +  β =   -b + √ b 2 - 4ac  +  - b - √ b2 -4ac   =   -2b  =  -b    = -coefficient of x
2a 2a  2a a coefficient of x2
Hence sum of the roots = -b/a for the quadratic equation ax2 + bx + c = 0

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Examples of Sum of the Roots Formula :-
Let us see some examples on sum of the roots:-
#Find the sum of the roots of the equation x2 + 7x + 6 = 0
Solution:-
Step 1 : This problem is of the form ax2 + bx + c = 0
Step 2 : Hence a = 1 and b = 7
Step 3 : Sum of the roots = -b/a = -7/1 = -7
Answer Sum of the roots of the equation x2 + 7x + 6 = 0 is -7
Here is another interesting problem.

Find the  value of k  so that the equation 49x2 - kx - 81 = 0 has one root as the negative of the other root.
Solution: -
Step 1 : In this equation a = 49  and b =-k
Step 2 : Let one root be α.  Then the other root is -α
Step 3 : Sum of the roots is α + (-α) =  - b / a = - (-k)/ 49
α  -  α = k / 49
0 = k / 49
49 * 0 = k
Hence value of k = 0
Answer  k = 0

Formula for Sum of Roots of a Complex Number :-
Let us see an example the sum of the roots of a complex  equation.
# One root of a complex equation is 2 - i.  Find the equation.
Solution:
Step 1 : One root of the equation is (2 - i)
Step 2 : Then the other root is its conjugate and it is ( 2 + i)
Step 3 : Then the equation is written as  [ x  - ( 2 - i ) ] [ x - ( 2 + i)]
=     [ x - 2 - i ][ x - 2 + i]
=     ( x - 2)2  - ( i2)  using   Formula  a2 - b2 = ( a -b )( a + b)
=      x2 - 4x + 4 - (-1)
=      x2   - 4x + 4 + 1
=      x2  - x + 5  = 0

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Problem Based on the Sum of the Roots of Unity:-
Let us find the sum of the fourth roots of unity:
Find the fourth roots of unity and  the sum.
Solution:-
Step 1 : Let us write the  problem as (1)1/4
Step 2 : Let x = (1)1/4
Step 3 : Then x4  =  (1)1/4 * 4 = > x 4 =  1
Step 4 : This gives us  x4 - 1 = 0
Step 5 : Let us factorise the equation .  We get (x2 - 1) (x2 + 1 ) = 0
(x -1) (x + 1) (x2 + 1) = 0
x = 1 ; x = -1 and (x2 + 1 ) = 0
Step 6 : Let us factorise (x2 + 1 ) = 0
x2  = -1
x2 = ± √ -1
x 2= ± √ i2
Hence x = ± i
Answer : Hence the four roots are +1, -1 ,+i and -i
Sum of the four roots are +1 +(-1) + i +(-i) = 1 - 1 + i - i = 0 + 0 = 0
Sum of the roots of unity s  zero.