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Math Help - Complex number roots

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by Murphie View Post
    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, b^2 - 4ac, for a quadratic equation in the form ax^2 + bx + c = 0.

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

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

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

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


    01
    Last edited by yeongil; July 5th 2009 at 05:28 AM.
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  3. #3
    MHF Contributor

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    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 (x-(a+ bi))(x-(a-bi))= x^2- (a+bi)x- (a-bi)x+ (a-bi)(a+bi) = x^2- 2ax+ a^2+ b^2= x^2- 4x+ 21.

    So -2a= -4 and a^2- b^2= 21. Obviously "-2a= -4" tells us that a= 2. Putting that into a^2- b^2= 21 we have 4+ b^2= 21 or 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 (x-(a+bi))(x- (c+di))= x^2- (a+bi)x- (c+di)x+ (a+bi)(c+di) = 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.
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