Expert: Sherman D. - 9/22/2004

Question
Hi Sherman!

I have a question concerning two assignments in my math books. The topic is real and imaginary roots for second degree and third degree polynomials.

The first one I think I have solved:

For the roots sqrt2 and PI, formulate a second degree equation with real coefficients:

This should be:

(x- x1) (x ¨C x2) = x2 ¨C x1*x- x2*x+x1* x2 = x2-
(x1+ x2)x+ x1* x2 =

x^2-( x1+ x2)x+ (x1* x2) = x^2 ¨C (pi+¡Ì2)x+(pi*¡Ì2)


x2- 4.555806216x + 4.442882938 = 0

The second assignment, however, I am unable to solve:

For the complex double root -i and the complex root i, formulate a third degree polynomial with complex coefficients:

Am I right in thinking that this should be:

x^3 - (i+(-i)+(-i)*x^2 + (i*(-i)+i*(-i)+(-i)*(-i)*x+(i*(-i)*(-i) = 0


x^3 - (-i) * x^2+(-i^2+ (-i^2)+( i^2)*x+(i*(i^2) = 0


x3 - (-i) * x2+(1)*x+(i*(-1) = 0


But now I unsure what to do?

Best regards,  

Answer
(x - sqrt(2))(x - PI) =
x^2 - PIx - sqrt(2)x + PIsqrt(2)
x^2 + (-PI - sqrt(2))x + PIsqrt(2)
x^2 - 4.555806215983x + 4.555806215983 = 0

by the way, sqrt stands for square root. Some people don't know that when i type that.

-----------------------------------------------------------

if by double root, you mean roots -i, -i, and complex root i, then

(x + i)(x + i)(x - i)

(x^2 + ix + ix + i^2)(x - i)
(x^2 + 2ix + i^2)(x - i)

Now you know that "i" is the sqrt of -1, and since this is being squared, the sqrt is canceled out, so this gives you

(x^2 + 2ix - 1)(x - i)

x^3 - ix^2 + 2ix^2 - 2i^2x - x + i

x^3 + (-1 + 2)ix^2 + (((-2)(-1)) - 1)x + i

x^3 + ix^2 + (2 - 1)x + i

x^3 + ix^2 + x + i

That is what i got. If you have any questions, just let me know.

-----------------------------------------------------------

also when dealing with creating an equation using the roots, the formula is

(ax - b)(cx - d)

where as "a" and "c" are the denominators, and "b" and "d" are the numerators.

when dealing with 3 roots, its basically the same thing

(ax - b)(cx - d)(ex - f)

and so on.