# What is the equation of f(x) if (3-5i) and (3+5i) are the roots?

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Let the equation be f(x) = x^2 + bx + c

Then we know that if x1 and x2 are the roots of f(x).

Then x1+x2= -b/a

==> x1*x2= c/a

We have the roots (3-5i) and (3+5i)

==> (3-5i)+(3+5i) = 6 = -b ==> b= -6

==> (3-5i)(3+5i)= 9+25= 34 = c

==> f(x)= x^2 - 6x + 34 = 0

To check we will calculate the roots.

==> x1= (6+sqrt(-100) / 2 = 6+10i / 2 = 3+ 5i

==> x2= 3-5i

**Then the equation is: f(x) = x^2 -6x + 34 **

The roots of the equation are 3 - 5i and 3 + 5i

We can write this as (x - (3 - 5i))(x -(3 + 5i)) = 0

=> (x - 3 + 5i)(x - 3 - 5i) = 0

=> (x - 3)^2 - (5i)^2 = 0

=> x^2 +9 - 6x - 25(-1) = 0

=> x^2 + 9 - 6x + 25 = 0

=> x^2 - 6x + 34 = 0

**The required equation for f(x) is x^2 - 6x + 34 = 0**