1. ## Complex number roots

Find two complex number roots for the equation x^2-4x+21=0
Not sure I know what this means, could someone explain it to me.

I tried to do this using synthetic division? But couldn't find the zero...
7| 1 -4 21
1 7 21

1 3 42

-3| 1 -4 21
1 -3 21

1 -7 42
I think i should have gotten a -21 resulting in a 0
For any help:
Thanks!

2. Originally Posted by Murphie
Find two complex number roots for the equation x^2-4x+21=0
Not sure I know what this means, could someone explain it to me.
If the roots are complex (in the form a + bi), then there are no real solutions. And we can show this by checking the discriminant, $\displaystyle b^2 - 4ac$, for a quadratic equation in the form $\displaystyle ax^2 + bx + c = 0$.

\displaystyle \begin{aligned} b^2 - 4ac &= (-4)^2 - 4(1)(21) \\ &= 16 - 84 \\ &= -68 \end{aligned}

Since the discriminant is negative, there are no real solutions.

We find the roots by using the quadratic formula:
\displaystyle \begin{aligned} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{4 \pm \sqrt{(-4)^2 - 4(1)(21)}}{2(1)} \\ &= \frac{4 \pm \sqrt{-68}}{2} \\ &= \frac{4 \pm 2i\sqrt{17}}{2} \end{aligned}

So
$\displaystyle x = 2 + i\sqrt{17}$ and
$\displaystyle x = 2 - i\sqrt{17}$.

01

3. Another way to do this:

Since the coefficients of the equation are real, the roots must be "complex conjugate"- of the form a+ bi and a- bi. Then we must have $\displaystyle (x-(a+ bi))(x-(a-bi))= x^2- (a+bi)x- (a-bi)x+ (a-bi)(a+bi)$$\displaystyle = x^2- 2ax+ a^2+ b^2= x^2- 4x+ 21. So -2a= -4 and \displaystyle a^2- b^2= 21. Obviously "-2a= -4" tells us that a= 2. Putting that into \displaystyle a^2- b^2= 21 we have \displaystyle 4+ b^2= 21 or \displaystyle b^2= 17, giving the answers yeongil gave.' How do we know that the roots must be "complex conjugate"? Suppose they were just the general a+bi and c+di. Then \displaystyle (x-(a+bi))(x- (c+di))= x^2- (a+bi)x- (c+di)x+ (a+bi)(c+di)$$\displaystyle = x^2- ((a+c)+ (b+d)i)x+ (ac- bd+ (ad+bc)i$. In order that the coefficients be real, the imaginary parts, b+d and ad+bc, must be 0. From b+ d= 0 we have d= -b and putting that into ad+ bc= 0 we have -ab+ bc= b(c-a)= 0. Either b= 0, in which case we have d= -b= 0 so the roots are real, or c-a= 0 so that c= a and then we have ad+ ab= 0 giveing a(d+ b)= 0. In that case, d= -b so the roots must be a+ bi and c+ di= a- bi, complex conujates.