Question

I really need help with a quadratic question.

So the problem is
kx^2+[k+2]x-3=0 has roots which are real and positive. Find the possible values that k may have.

The answer is -8+ √60<k<0
I get why they got -8+ √60 but I don't understand the 0. I thought it would be
-8+ √60 <k < -8- √60

I will very much appreciate it if anyone could help me with this problem!

Answer 1

using the quadratic formula, we have

x = [-(k+2)±√((k+2)^2 + 12k)]/2k

For the roots to be real, we need
(k+2)^2 + 12k >= 0
-8-√60 <= k <= -8+√60

Now, we want the roots to be positive, so we need

[-(k+2)±√((k+2)^2 + 12k)]/2k > 0

Answer 2

Tough question!

roots are:
x = ( -k-2 ± √(k+2)^2 - 4(k)(-3) )/2k
= ( -k-2 ±√(k^2 + 4k + 4 + 12k))/2k

it said: both x's are positive, so

( -k-2 + √(k^2 + 4k + 4 + 12k))/2k > 0
assuming we started with a positive x^2 term, so k > 0
-k-2 + √(k^2 + 4k + 4 + 12k) > 0
√(k^2 + 16k + 4) > k+2
square both sides:
k^2 + 16k + 4 > k^2 + 4k + 4
12k > -4
k > -1/3

or

-k-2 - √(k^2 + 4k + 4 + 12k) > 0
-k-2 > √(k^2 + 4k + 4 + 12k)
square both sides
k^2 + 4k + 4 > k^2 + 16k + 4
0 > 12k
k < 0

for real roots:
b^2 - 4ac > 0
(k+2)^2 -4k(-3) > 0
k^2 + 4k + 4 + 12k > 0
k^2 + 16k + 64> -4 + 64 <----- completed the square
(k+8)^2 > 60
±(k+8) > √60

k+8 > √60 , k > -8+√60 , ( k > appr -.25)
or
-k-8 > √60
-k > 8 + √60
k < -8 - √60 , (k < appr -15.75)

so to be real, -15.75

so testing (since we squared, all answers must be tested), we have 4 conditions
k < 0
k > -1/3
k > -8 + √60 or k > appr -.25
k < -8 - √60 or k < -15.75

and using Wolfram , we can see that
indeed -8+√60 < k < 0

http://www.wolframalpha.com/input/?i=solve+kx%5E2%2B%5Bk%2B2%5Dx-3%3D0+for+k%3D+-.2

change the value of k to something like k = 2 and you will see one positive and one negative root
try different values of k