Maths help core 1
Can someone please help me with this confusing math problem on discriminant and quadratic inequalities?
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?
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?
Original post by wednesday_adams
Can someone please help me with this confusing math problem on discriminant and quadratic inequalities?
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?
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
(edited 9 years ago)
Original post by kkboyk
Made a mistake with the sign 
...
...I don't understand? At what stage do I make it >=0?
Original post by wednesday_adams
Can someone please help me with this confusing math problem on discriminant and quadratic inequalities?
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
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
Posted from TSR Mobile
(edited 9 years ago)
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
Posted from TSR Mobile
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
Posted from TSR Mobile
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?
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
Posted from TSR Mobile
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
Posted from TSR Mobile
(edited 9 years ago)
Original post by wednesday_adams
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?
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
Original post by wednesday_adams
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?
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).
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).
(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
Original post by TenOfThem
NOTE that this is incorrect
Would you like to explain how
Original post by ttaylor17
Would you like to explain how
The correct working was given in Post 7
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
(edited 9 years ago)
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
Posted from TSR Mobile
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
Posted from TSR Mobile
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