# Maths help core 1Watch

#1
I think I'm doing it all wrong...

It says:

Find the range or ranges of values k can take for kx^2-4x+5-k=0 to have real roots.

So I did this:

a=k
b=-4
c=5-k

(-4^2) -4(k)(5-k)>= (greater than or equal to) 0

16-20k+4k^2>=o

Then I subtracted 16.
4k^2-20k>=-16

then I didn't know what to do! i thought maybe I could factor the 4 out like this:
4k(k-5)>=-16

or should I just divide everything by 4 like this:
K^2-5k>=-4
and then factor the k out like this
k(k-5)>=-4

But then I'm stuck with what to do from here.
If I added 5 to both sides I would get
k(k)>=1
and then would I factor the k back in and get
k^2>=1
k>= -1 or =1?

Would that be the answer or am I completely way off?
0
4 years ago
#2
Are you familiar with quadratic inequalities at all?

Posted from TSR Mobile
0
4 years ago
#3
I think I'm doing it all wrong...

It says:

Find the range or ranges of values k can take for kx^2-4x+5-k=0 to have real roots.

So I did this:

a=k
b=-4
c=5-k

(-4^2) -4(k)(5-k)>= (greater than or equal to) 0

16-20k+4k^2>=o

Then I subtracted 16.
4k^2-20k>=-16

then I didn't know what to do! i thought maybe I could factor the 4 out like this:
4k(k-5)>=-16

or should I just divide everything by 4 like this:
K^2-5k>=-4
and then factor the k out like this
k(k-5)>=-4

But then I'm stuck with what to do from here.
If I added 5 to both sides I would get
k(k)>=1
and then would I factor the k back in and get
k^2>=1
k>= -1 or =1?

Would that be the answer or am I completely way off?
You should certainly take the factor 4 out to make things simpler.

But when you get k(k -5 ) >= -4, multiply out and rearrange to get

k^2 -5k + 4 >= 0

This is a standard quadratic inequality which you should be able to factorize by inspection. Have you seen things like this before?
0
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