# 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
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?
First of all make it >0 and divide by 4 then factorise. You should get (K-1)(K-4)<0
0
4 years ago
#3
Made a mistake with the sign ...
0
#4
(Original post by kkboyk)
Made a mistake with the sign ...
I don't understand? At what stage do I make it >=0?
0
4 years ago
#5
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
That final step is where you went wrong
You have a quadratic and you'd never solve a quadratic in the manner you described.
4k^2 - 20k + 16 >= 0
Divide through by 4 to simplify the quadratic
From the result use your quadratic skills to find the critical values where it = 0
Use the solutions to sketch the curve of the inequality and then note the values of k where the curve is >= 0

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0
#6
(Original post by gdunne42)
That final step is where you went wrong
You have a quadratic and you'd never solve a quadratic in the manner you described.
4k^2 - 20k + 16 >= 0
Divide through by 4 to simplify the quadratic
From the result use your quadratic skills to find the critical values where it = 0
Use the solutions to sketch the curve of the inequality and then note the values of k where the curve is >= 0

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Thanks, I understand now.

I factored it and I got (k-1)(k-4)>=0
Do I have to sketch the curve in order to get the answer or is the answer k=-1 or k=-4?
0
4 years ago
#7
You are solving an inequality that is >= 0
So no, those are not the final answers
Sketching the quadratic of k isn't required but it makes it easy to see what range of values of k satisfy the inequality
The 'critical values' for the sketch are k= 1 or k=4 which is where it crosses the axis (=0)
From the sketch you can see the curve is >= 0 if k <=1 or k>=4

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0
4 years ago
#8
Thanks, I understand now.

I factored it and I got (k-1)(k-4)>=0
Do I have to sketch the curve in order to get the answer or is the answer k=-1 or k=-4?
Shouldn't it be k = 4 and 1 instead of -4 and -1
0
4 years ago
#9
Thanks, I understand now.

I factored it and I got (k-1)(k-4)>=0
Do I have to sketch the curve in order to get the answer or is the answer k=-1 or k=-4?
(k-1)(k-4)>=0 can be written as:

(k-1)>=0
and
(k-4)>=0

Therefore, if you solve these you get:
k>=1 and k>=4 (and not the answers you gave).

It's good practice to now you draw a U shaped graph (positive x squared graph) and mark point above the x axis (in this case the k axis).
0
4 years ago
#10
(Original post by borysek01)
(k-1)(k-4)>=0 can be written as:

(k-1)>=0
and
(k-4)>=0

Therefore, if you solve these you get:
k>=1 and k>=4
(and not the answers you gave).

It's good practice to now you draw a U shaped graph (positive x squared graph) and mark point above the x axis (in this case the k axis).
NOTE that this is incorrect
0
4 years ago
#11
(Original post by TenOfThem)
NOTE that this is incorrect
Would you like to explain how
0
4 years ago
#12
(Original post by ttaylor17)
Would you like to explain how
The correct working was given in Post 7
0
4 years ago
#13
(Original post by ttaylor17)
Would you like to explain how
for example k=-1 does not satisfy the conditions you state for k but if you substitute it into (k-1)(k-4) the result is > 0
0
4 years ago
#14
To have real roots b2-4ac > 0
To have one repeated its =0

So when you have your k^2+xk>c
Make that into a quadratic and get factors i.e. K^2 + xk -c > 0

I added x and c as i forgot what these were

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