How to solve (corrected)?

Question

How to solve?

i would like to know the steps to solve this:

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


tanks

Answer 1

Easy.

Write f(x) = (x-a)(x-b)
where a and b are the 2 solutions.
Let a = 3b
f(x) = (x-3b)(x-b)
= x^2 - 4bx + 3b^2

b = 2
and k = -4b = -8
The other root is 6.

Answer 2

The roots are [-k ± sqrt(k²-48)] / 2. So our equation is

[-k ± sqrt(k²-48)] / 2 = 3 * [-k -/+ sqrt(k²-48)] / 2

→ -k ± sqrt(k²-48) = -3k -/+ 3sqrt(k²-48)

→ 2k = ± 4sqrt(k²-48)

→ k = ± 2sqrt(k² - 48)

→ k² = ± 4(k² - 48)

→ k = -8 (other solutions don't satisfy either k<0 or a positive discriminant).

So the quad is x² - 8x + 12 = (x-2)(x-6) and the root 6 is three times the other root 2.

Answer 3

Let the roots be P and Q.

PQ = 12
P=3Q
(3Q)Q=12
3Q^2=12
Q^2=4
Q= +2 or -2


P+Q = k<0, so must be Q=-2, P=-6
P+Q = k = -8

Answer 4

a=root 1
3a=root 2
The product of the roots has to equal 12: 3a^2=12
so a^2=4
so a=-2 and k would equal -4.

Answer 5

ouch that looks hard