Finding quadratic equation (complex numbers)

TsAmE · **Joined:** Mon, 19 Apr 2010 13:45:05 UTC **Posts:** 61

Find a quadratic equation with real coefficients that has 3 + i as one of its roots.

I am not sure what to

Matt · **Joined:** Wed, 1 Oct 2003 04:45:43 UTC **Posts:** 9633

If 3 + i is a root, then 3 - i has to be also.

f(x)=[x-(3+i)][x-(3-i)]

Sam Snyder · **Posted:** Tue, 19 Oct 2010 15:57:33 UTC

TsAmE wrote:

Find a quadratic equation with real coefficients that has 3 + i as one of its roots.

I am not sure what to

Method 1:
Complex solutions to polynomial equations (with real coefficients) occur in complex conjugate pairs. Therefore, 3-i

is also a root of this equation.
Multuply $[x-(3+i)]\cdot[x-(3-i)]$ and set this equal to zero.
Regrouping may help you do this. $[(x-3)-i]\cdot[(x-3)+i]$

(Can you see that this will give you a difference of squares?)
.
Method 2:
The solution to the quadratic equation, ax^2+bx+c=0

is $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ .
Taking a=1

will simplify your task.
You know that one of the roots of the quadratic is 3+i

, so $$3+i=\frac{-b+\sqrt{b^2-4c}}{2}$ .
The imaginary part of the root must come from taking the square root of a negative number. The real part is -b/2

, so: $$-\frac{b}{2}=3$ .

Now that you know b, set $$\frac{\sqrt{b^2-4c}}{2}=i$ .

Square both sides & solve for c.

Hopefully, you can take it from there to get your answer.

S.O.S. Mathematics CyberBoard

Finding quadratic equation (complex numbers)

Who is online