# f(x) = x^2 + mx - 15 if 3 is one of the root, find m and the second root if possible.

*print*Print*list*Cite

f(x) = x^2 + mx -15

Since x=3 is one of the roots for f(x) , then f(3) =0

Let us substitute:

f(3) = 3^2 + m*3 - 15 = 0

==> 9 +3m -15 = 0

==> 3m -6 = 0

==> 3m = 6

==> m = 6/3 = 2

==> **m =2**

Then:

f(x) = x^2 +2x - 15

let us factor:

f(x) = (x-3)(x+5)

Then, the roots are:

x1= 3

**x2= -5**

We'll use Viete's relations to calculate the other root of f(x) = x^2 + mx - 15:

x1 + x2 = -b/a

x1*x2 = c/a

We'll identify a,b,c:

a = 1

b = m

c = -15

x1 + x2 = -m

But x1 = 3

3 + x2 = -m (1)

x1*x2 = -15

3*x2 = -15

x2 = -5

We'll substitute x2 in (1):

3 - 5 = -m

**m = 2**

f(x) = x^2+mx-15.

Given 3 is one root. To find m.

Solution:

A root a1( or zero) of the f(x) makes the value f(a1) = 0.

So a1 =3. Therefore f(3) = 3^2+m*3-15 = 0

9+3m-15 =0

3m -6 = 0

3m = 6

3m/3 = 6/3

m = 2

To find the other root:

So the function f(x) becomes x^2 +2x -15.

Factorise x^2+2x-15:

x^2+5x-3x -15

x(x+5) -3(x+5)

(x+5)(x-3)

So f(x) = 0 gives (x+5)(x-3) = 0

So x+5 = 0 or x-3 = 0

x+5 = 0 gives **x = -5 as the other root**.

x-3 = 0 gives x = 3 the root given by data.