answersLogoWhite

0

Hi,

I'm not sure where you got this question, but it can't be factorised (using real numbers) as it has no real roots.

The only solutions to this question are imaginary. (Don't worry about imaginary numbers unless you are in high school and doing the equivalent of ext. 2 maths in the NSW HSC (Yr 12 exam) in Australia or equivalent).

(If you need the imaginary roots see below)

To find out if a question has real roots, use the equation;

b^2 - 4ac.

If the equation has real solutions, then this must give a positive number.

This only applies to quadratic equations. (ax^2 + bx +c = 0).

e.g. For a^2 + 10a + 30,

a is the pronumeral equivalent to x...

Thus the equation is;

1(a^2) + 10(a) + 30 = 0

and b^2 - 4ac equals 10^2 - 4*1*30 which then gives 100 - 120 = -20.

As the number is negative, there are no real roots. (and thus can't be factorised with only real numbers.)

If you are using imaginary numbers, look at this.

x = (-b +/- (sqrt(b^2 - 4ac)))/2a (Note: '+/-' is plus or minus indicating one

root is found by adding and the other is

found by subtracting)

Thus;

a = (-10 +/- (sqrt(-20)))/2 ( -20 is from above working)

a = (-10 +/- i(sqrt(20)))/2

a = -5 +/- i(sqrt(5))

These are the solutions to the question and as such the roots are;

(a + 5 - i(sqrt(5))) (a + 5 + (sqrt(5)))

Hope this helped...

jdru4706

User Avatar

Wiki User

16y ago

What else can I help you with?