Question

The roots of a quadratic equation ax2+bx+c=0 are 1+i 5 and 1−i 5 . Find possible values of a, b and c.

Answer 1

Answer:

Step-by-step explanation:

If the roots are 1 + 5i and 1 - 5i, then you need the factors that result from those roots.  They are (x - 1 + 5i) and (x - 1 - 5i).  Now what you do with those is FOIL them out.  Doing that gives you the following:

$x^2-x+5ix-x+1-5i-5ix+5i-25i^2$ (what a mess, huh?)

The good thing is that several of those terms cancel each other out.  +5ix cancels out the -5ix; -5i cancels out the 5i; and the 2 -x terms combine to -2x.  That leaves you with:

$x^2-2x-25i^2$

Obviously you're in the section in math that deals with complex (imaginary) numbers so you should know that i-squared is equal to -1.  Making that replacement:

$x^2-2x+25$

a = 1, b = -2, c = 25

Answer 2

Answer:

$a\neq 0$, $b=-2a, c=6a$

Step-by-step explanation:

Use Vieta's theorem:

$\frac{-b}{a} = 1+i\sqrt{5}+1-i\sqrt{5} = 2$

so b=-2a

$\frac{c}{a} = (1+i\sqrt{5})(1-i\sqrt{5} ) = 1+5 = 6$

so c = 6a

Coefficient a can technically be any number except 0, because if we look at the vertex form of the ax2+bx+c=0 equation, which is: a(x-r)(x-v)=0 (r and v being the roots), we can see that any value of a would end up with the same resulting quadratic. (we're just multiplying the whole equation by one number).

However, a can't equal zero because if we set a as zero in ax^2+bx+c=0, this equation wouldn't be a parabola anymore. There would be no x^2, and the the graph would just be a linear line.

So if we combine the two cases of a being any number and it not being zero, the result is $a\neq 0$.