How to solve?

Question

i would like to know the steps to solve this:

One root of f(x)=x^2+12 is 3 times the other root. Find the roots and the value of k if k<0.

tanks

it's f(x) =x^2+kx+12 is 3 **

Answer 1

f(x)=x^2+kx+12

f(x)=(x+k1)(x+k2) and k1=3k2

x^2+(k1+k2)x+(k1)(k2)

(k1)(k2)=12
3(k2)(k2)=3(k2)^2=12 --> k2=2, k1=6

So the roots are x+k1=0 and x+k2=0 so the roots are x=-2 and x=-6 and k=-8

Answer 2

Did you mean x² + 12x + k?
It seem's you left out something here!
I'll assume that's what you meant.
Let r and s be the roots.
Then s = 3r and k = 3r², so k can't be negative unless
the roots are complex. But
r + s = -12
r + 3r = -12
r = -3
s = -9
k = 27.
Check the roots of x² + 12x + 27 = 0
are -9 and -3.
Let's do the same question for x² + kx + 12.
Now let the roots be r and s
Then rs = 12
r+s = -k
But s = 3r
So 3r² = 12
and we must take r = -2 to make k negative
Then s = -6.
k = -8.
Check: The roots of x² -8x + 12 = 0 are 6 and 2.

Answer 3

x^2 + 12 does not have real roots.

the roots are in fact i sqrt(12) and -i sqrt(12). they necessarily are conjugates and have the same amplitude so you can never have one root 3 times the other with this shape of equation.

Answer 4

k does not exist.

check the equation

correct your question.

Answer 5

There is no k in your question

Answer 6

the answer is obvious, k= -24 x c squared, duh.