# How do we get k<-1, k>2 from (k+1)(k-2)>0

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#1
I have to find the range of k for which each quadratic equation has two different positive solutions.

We are given the quadratic equation Let f(x) = x^2-2kx+k+2

Using the Discrminant D/4 = b^2-ac,
How do we get k<-1, k>2 from (k+1)(k-2)>0??

I just can't seem to work that out???
0
9 months ago
#2
Note the condition using the discriminant for two distinct roots is:

Then as you have pointed out, you obtain equation:

Notice this is a quadratic which is concave up / smiley face.
For which value of k is the equation * true (simple to read off from equation * as it is already factorised) ?
Last edited by simon0; 9 months ago
1
#3
(Original post by simon0)
Note the condition using the discriminant for two distinct roots is:

Then as you have pointed out, you obtain equation:

Notice this is a quadratic which is concave up / smiley face.
For which value of k is the equation * true (simple to read off from equation * as it is already factorised) ?
That is really appreciated. 3 people couldn't explain it to me as simple as you have.
Cheers
1
9 months ago
#4
(Original post by SunnyCoast)
I have to find the range of k for which each quadratic equation has two different positive solutions.

We are given the quadratic equation Let f(x) = x^2-2kx+k+2

Using the Discrminant D/4 = b^2-ac,
How do we get k<-1, k>2 from (k+1)(k-2)>0??

I just can't seem to work that out???
Are you after two positive solutions (as well as real)?
If so, k<-1 won't work and ypu'd have to justify the k>2 ones
1
9 months ago
#5
(Original post by SunnyCoast)
That is really appreciated. 3 people couldn't explain it to me as simple as you have.
Cheers
No problem.

Also as "mqb2766" has stated, the question asks for positive solutions, so you need to justify why the set of solutions of form:

are suitable.
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