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Finding values of k for two distinct real roots

Find values of k for two distinct real roots
y= kx^2 + (2k+1)x + (k+1)

The Answer is : The discriminant is 1. This means that as 1>0, the curve has two distinct real roots for all values of k.

I do not understand the bit which says "This means that as 1>0, the curve has two distinct real roots for all values of k"

What makes it confusing is that k has completely disappeared. Could someone help me with this please?

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Reply 1
Original post by sienna2266
Find values of k for two distinct real roots
y= kx^2 + (2k+1)x + (k+1)

The Answer is : The discriminant is 1. This means that as 1>0, the curve has two distinct real roots for all values of k.

I do not understand the bit which says "This means that as 1>0, the curve has two distinct real roots for all values of k"

What makes it confusing is that k has completely disappeared. Could someone help me with this please?

Have you calculated the discriminant using the usual
b^2-4ac

There is a slight error in the solution k=0 does not produce 2 roots.
Reply 2
Original post by mqb2766
Have you calculated the discriminant using the usual
b^2-4ac

There is a slight error in the solution k=0 does not produce 2 roots.

Yes, this is how I went about it.
a=k
b= 2k+1
c =k+1
(2k+1)^2 -4(k)(k+1)>0
4k^2 + 4k +1 -4k(k+1)>0
Which resukts in 1>0.

Dont understand the answer though.
Kindly let me know what you think about this please.
Many thanks
Reply 3
Original post by sienna2266
Yes, this is how I went about it.
a=k
b= 2k+1
c =k+1
(2k+1)^2 -4(k)(k+1)>0
4k^2 + 4k +1 -4k(k+1)>0
Which resukts in 1>0.

Dont understand the answer though.
Kindly let me know what you think about this please.
Many thanks


You must realise the discrrimiant comes from the quadratic equation so
x = -b/2a +/- sqrt(b^2-4ac)/2a
so -b/2a is the average value of the two roots (if they exist) and sqrt(b^2-4ac)/2a is how much each root is either side of the average. So if
b^2-4ac > 0
then the two roots are distinct, as one is greater than the average and one is less than the average.

Youve shown this is the case as
b^2-4ac = 1
and 1 is always > 0. It doesnt depend on k (nearly) but the location of the roots do
https://www.desmos.com/calculator/wnrowiaxjn
as both -b/2a and +/-1/2a do depend on k.

Note that the quadratic coefficient is k, so if this is zero, then the function is a linear function and there is only a single root. So the discriminant analysis is invalid in this special case. So really there are two distinct roots for all k!=0.
(edited 7 months ago)
Reply 4
Original post by sienna2266
Find values of k for two distinct real roots
y= kx^2 + (2k+1)x + (k+1)

The Answer is : The discriminant is 1. This means that as 1>0, the curve has two distinct real roots for all values of k.

I do not understand the bit which says "This means that as 1>0, the curve has two distinct real roots for all values of k"

What makes it confusing is that k has completely disappeared. Could someone help me with this please?

Here is some content:

--> the bit under the square root in the quadratic formula is b^2 - 4ac.
1) sq rooting a positive value will give two solutions, i.e., +/- x
2) sq rooting a negative value will not yield any real solutions (in other words, you can't sq root a negative number)
3) sq rooting 0 gives +/- 0

Okay, now if you determine the discriminant using b^2 - 4ac, and you know that in the quadratic formula (to calculate the solutions), you will square root your discriminant: based on 1,2 and 3 above, you will get two, no real or one repeated solution respectively.

If you understand this then the question:
Find values of k for two distinct real roots, y= kx^2 + (2k+1)x + (k+1)
... is asking you to determine for what values of k will this quadratic have two real roots - so you would say, well, this will have two distinct roots when case (1) is true - in other words, when b^2 - 4ac > 0.

As you state in the solution, the k disappears, and b^2 - 4ac simplifies down to 1 > 0. How do you read this solution now? As long as the condition 1 > 0 is met, your initial statement (in bold) is true. Your initial statement was that the curve will have two distinct roots. 1 will always be greater than 0, and so your initial statement will always be true - the curve will always have two distinct roots - in other words, the curve will have two distinct roots regardless of what k is!

Note: as post #4 notes, k != 0

I've tried to explain the answer, I hope this is within the rules of the forum.
Reply 5
Original post by mqb2766
You must realise the discrrimiant comes from the quadratic equation so
x = -b/2a +/- sqrt(b^2-4ac)/2a
so -b/2a is the average value of the two roots (if they exist) and sqrt(b^2-4ac)/2a is how much each root is either side of the average. So if
b^2-4ac > 0
then the two roots are distinct, as one is greater than the average and one is less than the average.

Youve shown this is the case as
b^2-4ac = 1
and 1 is always > 0. It doesnt depend on k (nearly) but the location of the roots do
https://www.desmos.com/calculator/wnrowiaxjn
as both -b/2a and +/-1/2a do depend on k.

Note that the quadratic coefficient is k, so if this is zero, then the function is a linear function and there is only a single root. So the discriminant analysis is invalid in this special case. So really there are two distinct roots for all k!=0.

Many thanks for your help.

So does that mean that the discriminant is not always accurate when trying to find out if a quadratic has two real roots because the statement "the curve will have two distinct real roots regardless of what k is" doesn't stand when k cant be equal to zero?

didnt let me like:/ said rate other members for some reason
Reply 6
Original post by tamilGuy.
Here is some content:

--> the bit under the square root in the quadratic formula is b^2 - 4ac.
1) sq rooting a positive value will give two solutions, i.e., +/- x
2) sq rooting a negative value will not yield any real solutions (in other words, you can't sq root a negative number)
3) sq rooting 0 gives +/- 0

Okay, now if you determine the discriminant using b^2 - 4ac, and you know that in the quadratic formula (to calculate the solutions), you will square root your discriminant: based on 1,2 and 3 above, you will get two, no real or one repeated solution respectively.

If you understand this then the question:
Find values of k for two distinct real roots, y= kx^2 + (2k+1)x + (k+1)
... is asking you to determine for what values of k will this quadratic have two real roots - so you would say, well, this will have two distinct roots when case (1) is true - in other words, when b^2 - 4ac > 0.

As you state in the solution, the k disappears, and b^2 - 4ac simplifies down to 1 > 0. How do you read this solution now? As long as the condition 1 > 0 is met, your initial statement (in bold) is true. Your initial statement was that the curve will have two distinct roots. 1 will always be greater than 0, and so your initial statement will always be true - the curve will always have two distinct roots - in other words, the curve will have two distinct roots regardless of what k is!

Note: as post #4 notes, k != 0

I've tried to explain the answer, I hope this is within the rules of the forum.

Many thanks for your help. Also helped clear up things :smile:

So does that mean that the discriminant is not always accurate when trying to find out if a quadratic has two real roots because the statement "the curve will have two distinct real roots regardless of what k is" doesn't stand when k cant be equal to zero?
Reply 7
The discriminant is used to determine whether a quadratic has two distinct roots.
However, when k=0, its not a quadratic, so trying to use a discriminant (note it corresponds to a division by 0) doesnt make sense.

Original post by sienna2266
Many thanks for your help. Also helped clear up things :smile:

So does that mean that the discriminant is not always accurate when trying to find out if a quadratic has two real roots because the statement "the curve will have two distinct real roots regardless of what k is" doesn't stand when k cant be equal to zero?
Reply 8
Original post by mqb2766
The discriminant is used to determine whether a quadratic has two distinct roots.
However, when k=0, its not a quadratic, so trying to use a discriminant (note it corresponds to a division by 0) doesnt make sense.

Ah I see. Many thanks :smile:
Sorry to bother but just got slightly confused when you said "note it corresponds to a division by 0". What did you mean here?
Reply 9
Original post by sienna2266
Ah I see. Many thanks :smile:
Sorry to bother but just got slightly confused when you said "note it corresponds to a division by 0". What did you mean here?

The quadratic equation for the roots is
x = -b +/- sqrt(b^2-4ac) / 2a
So if a=0, then youre dividing by 2a = 0 which is undefined. A quadratic must have a non zero quadratic coefficient.
Reply 10
Original post by mqb2766
The quadratic equation for the roots is
x = -b +/- sqrt(b^2-4ac) / 2a
So if a=0, then youre dividing by 2a = 0 which is undefined. A quadratic must have a non zero quadratic coefficient.

Many thanks - this makes sense now :smile:
Original post by sienna2266
Many thanks - this makes sense now :smile:

I know it does, but just to add, the question should really have stated when defining the quadratic that k!=0. The moment k = 0, y= kx^2 + (2k+1)x + (k+1) becomes y = 0x^2 + x + 1, which is not a quadratic anymore. So for it to be a quadratic, k cannot be 0. Just pay attention to 'exceptions' like this one, though I would have expected the question to have said that. Goodluck.
Reply 12
Original post by tamilGuy.
I know it does, but just to add, the question should really have stated when defining the quadratic that k!=0. The moment k = 0, y= kx^2 + (2k+1)x + (k+1) becomes y = 0x^2 + x + 1, which is not a quadratic anymore. So for it to be a quadratic, k cannot be 0. Just pay attention to 'exceptions' like this one, though I would have expected the question to have said that. Goodluck.

In general, questions like this do not make such statements. Its up to the student to make sure they make clear/understand any assumptions when they perform the analysis. So if they perform quadratic analysis, they should check its a quadratic in the first place.
Reply 13
Original post by tamilGuy.
I know it does, but just to add, the question should really have stated when defining the quadratic that k!=0. The moment k = 0, y= kx^2 + (2k+1)x + (k+1) becomes y = 0x^2 + x + 1, which is not a quadratic anymore. So for it to be a quadratic, k cannot be 0. Just pay attention to 'exceptions' like this one, though I would have expected the question to have said that. Goodluck.

Many thanks :smile:
Just wondered if I could ask another question here on the topic of discriminant - no worries if you're busy.
On the attached file, you can see there's a graph for Question One. Is that the graph of the discriminant (y=k^2 - 3)? So do we have a k axis (y=0) and a y axis (k=0)?
So essentially, the question is asking: where is the graph of the discriminant (y=k^2 - 3) below the k axis ( y=0).
And the answer is it is between -√3 and √3.

Does what I've said above make sense or is my interpretation wrong?
Reply 14
Original post by mqb2766
In general, questions like this do not make such statements. Its up to the student to make sure they make clear/understand any assumptions when they perform the analysis. So if they perform quadratic analysis, they should check its a quadratic in the first place.


Original post by tamilGuy.
I know it does, but just to add, the question should really have stated when defining the quadratic that k!=0. The moment k = 0, y= kx^2 + (2k+1)x + (k+1) becomes y = 0x^2 + x + 1, which is not a quadratic anymore. So for it to be a quadratic, k cannot be 0. Just pay attention to 'exceptions' like this one, though I would have expected the question to have said that. Goodluck.

Many thanks :smile:
Just wondered if I could ask another question here on the topic of discriminant - no worries if you're busy.
On the attached file, you can see there's a graph for Question One. Is that the graph of the discriminant (y=k^2 - 3)? So do we have a k axis (y=0) and a y axis (k=0)?
So essentially, the question is asking: where is the graph of the discriminant (y=k^2 - 3) below the k axis ( y=0).
And the answer is it is between -√3 and √3.

Does what I've said above make sense or is my interpretation wrong?
Reply 15
Original post by sienna2266
Many thanks :smile:
Just wondered if I could ask another question here on the topic of discriminant - no worries if you're busy.
On the attached file, you can see there's a graph for Question One. Is that the graph of the discriminant (y=k^2 - 3)? So do we have a k axis (y=0) and a y axis (k=0)?
So essentially, the question is asking: where is the graph of the discriminant (y=k^2 - 3) below the k axis ( y=0).
And the answer is it is between -√3 and √3.

Does what I've said above make sense or is my interpretation wrong?

Looks good and yes that would be the graph of discriminant (y axis) verses k (x axis). You can always sub the answer(s) back into the original equation to check. Centre case is k=0 so the equation is
x^2 = -3
which obv. has no real solutions. Edge cases k=+/-sqrt(3) gives
x^2 +/-2sqrt(3)x + 3 = 0
which corresponds to the factorisation
(x+/-sqrt(3))^2 = 0
So they have a single root (discrimiant=0) which is what youd expect.

Note its geerally better to try and answer questions yourself rather than interpreting someone elses answer.
(edited 7 months ago)
Reply 16
Original post by mqb2766
Looks good and yes that would be the graph of discriminant (y axis) verses k (x axis). You can always sub the answer(s) back into the original equation to check. Centre case is k=0 so the equation is
x^2 = -3
which obv. has no real solutions. Edge cases k=+/-sqrt(3) gives
x^2 +/-2sqrt(3)x + 3 = 0
which corresponds to the factorisation
(x+/-sqrt(3))^2 = 0
So they have a single root (discrimiant=0) which is what youd expect.

Note its geerally better to try and answer questions yourself rather than interpreting someone elses answer.

Thanks so much for your time and my apologies for the delayed response - didn't get the chance to look at it properly.

I am confused about the centre case and edge cases you've mentioned. Not sure what they mean and why they are being used here, particularly k=0 :/ Also, why are we substituting into the original equation instead of the equation for the graph of the discriminant. A bit lost on what we are trying to achieve, but would really like to know what you did here.

Yeah that's a good idea. I mean I did A level maths 5 years ago but have forgotten everything and have lost all my notes including the past papers :frown:, so just brushing up on stuff first and then gonna have a go at some practise questions.
(edited 7 months ago)
Reply 17
Original post by sienna2266
Thanks so much for your time and my apologies for the delayed response - didn't get the chance to look at it properly.

I am confused about the centre case and edge cases you've mentioned. Not sure what they mean and why they are being used here, particularly k=0 :/ Also, why are we substituting into the original equation instead of the equation for the graph of the discriminant. A bit lost on what we are trying to achieve, but would really like to know what you did here.

Yeah that's a good idea. I mean I did A level maths 5 years ago but have forgotten everything, so just brushing up on stuff first and then gonna have a go at some practise questions.


It was more a suggestion to check "your" solution, though it seems like the solutions youre asking about are not your own. If you used the discriminant to say there are no real solutions to the original equation when
-sqrt(3) < k < sqrt(3)
you can check this answer by taking
k = 0, -sqrt(3), sqrt(3)
and subbing into the original equation. k=0 is particularly simple as the original equation becomes
x^2 = -3
which trivially has no real solutions as a squared term cannot be negative. Similarly the +/-sqrt(3) cases correspond to a repeated root which is what youd expect so -sqrt(3) < k < sqrt(3) would indeed be the correct interval for no roots.

If youre reviewing a level maths after 5 years, it may be worth going through worked textbook examples carefully, as theyll probably be more thoroughly explained than the solutions youre using here? Its boring advice to say read a textbook, but ...
(edited 7 months ago)
Reply 18
Original post by mqb2766
It was more a suggestion to check "your" solution, though it seems like the solutions youre asking about are not your own. If you used the discriminant to say there are no real solutions to the original equation when
-sqrt(3) < k < sqrt(3)
you can check this answer by taking
k = 0, -sqrt(3), sqrt(3)
and subbing into the original equation. k=0 is particularly simple as the original equation becomes
x^2 = -3
which trivially has no real solutions as a squared term cannot be negative. Similarly the +/-sqrt(3) cases correspond to a repeated root which is what youd expect so -sqrt(3) < k < sqrt(3) would indeed be the correct interval for no roots.

If youre reviewing a level maths after 5 years, it may be worth going through worked textbook examples carefully, as theyll probably be more thoroughly explained than the solutions youre using here? Its boring advice to say read a textbook, but ...

Thanks so much for your help. Didn't understand it at first so took a break from it and had a look at it again and all makes sense now.

When you substitute k= -√3 into y = x^2 + 2kx +3, you get y= x^2 - (2√3)x + 3.
The equation y= x^2 - (2√3)x + 3 has one real repeated root and that is x = √3. This is because when you solve for x (so sub y = 0 into y= x^2 - (2√3)x + 3), you get x= √3.
So the graph of y= x^2 - (2√3)x + 3 touches the x axis at x = √3.

When you substitute k= √3 into y = x^2 + 2kx +3, you get y= x^2 + (2√3)x + 3.
The equation y= x^2 + (2√3)x + 3 has one real repeated root and that is x = -√3. This is because when you solve for x (so sub y = 0 into y= x^2 + (2√3)x + 3), you get x= √3.
So the graph of y= x^2 + (2√3)x + 3 touches the x axis at x = -√3.

So you can conclude that when k = -√3 or k = √3, y = x^2 + 2kx +3 will have one repeated root.

When you substitute a value of k that is between -√3 and √3 (so when -√3 <k< √3) , you should get an error when solving for x.
For example, when you substitute k = 0 into y = x^2 + 2kx +3, you get y= x^2 + 3.
The equation y= x^2 + 3 has no real roots as it does not cross the x axis (y = 0).This is because when you solve for x (so sub y = 0 into y= x^2 + 3), you get x^2 = -3. You cannot square root a negative number, so you don't get a value for x when y = 0.
So the graph of = x^2 + 2kx +3 is above the x axis (since x^2 is positive - anything you square will give a positive value) when k = 0, it does not intersect the x axis.

You could try another value of k that is between -√3 and √3. When we sub k = 1 into y = x^2 + 2kx +3 and use the quadratic formula to solve for x, we get x = (-2 + √-8)/2 or x = (-2 - √-8)/2. This generates an error because we cannot square root a negative number, in this case -8.

So you can conclude that when -√3 <k< √3, y = x^2 + 2kx +3 will have no real roots.

:smile:
Reply 19
Original post by sienna2266
Thanks so much for your help. Didn't understand it at first so took a break from it and had a look at it again and all makes sense now.

When you substitute k= -√3 into y = x^2 + 2kx +3, you get y= x^2 - (2√3)x + 3.
The equation y= x^2 - (2√3)x + 3 has one real repeated root and that is x = √3. This is because when you solve for x (so sub y = 0 into y= x^2 - (2√3)x + 3), you get x= √3.
So the graph of y= x^2 - (2√3)x + 3 touches the x axis at x = √3.

When you substitute k= √3 into y = x^2 + 2kx +3, you get y= x^2 + (2√3)x + 3.
The equation y= x^2 + (2√3)x + 3 has one real repeated root and that is x = -√3. This is because when you solve for x (so sub y = 0 into y= x^2 + (2√3)x + 3), you get x= √3.
So the graph of y= x^2 + (2√3)x + 3 touches the x axis at x = -√3.

So you can conclude that when k = -√3 or k = √3, y = x^2 + 2kx +3 will have one repeated root.

When you substitute a value of k that is between -√3 and √3 (so when -√3 <k< √3) , you should get an error when solving for x.
For example, when you substitute k = 0 into y = x^2 + 2kx +3, you get y= x^2 + 3.
The equation y= x^2 + 3 has no real roots as it does not cross the x axis (y = 0).This is because when you solve for x (so sub y = 0 into y= x^2 + 3), you get x^2 = -3. You cannot square root a negative number, so you don't get a value for x when y = 0.
So the graph of = x^2 + 2kx +3 is above the x axis (since x^2 is positive - anything you square will give a positive value) when k = 0, it does not intersect the x axis.

You could try another value of k that is between -√3 and √3. When we sub k = 1 into y = x^2 + 2kx +3 and use the quadratic formula to solve for x, we get x = (-2 + √-8)/2 or x = (-2 - √-8)/2. This generates an error because we cannot square root a negative number, in this case -8.

So you can conclude that when -√3 <k< √3, y = x^2 + 2kx +3 will have no real roots.

:smile:


Sounds good. This is why I think its important to try and validate/understand the problem/solution, rather than just chugging through the algebra of someone elses answer
https://www.desmos.com/calculator/qqzldzolm1

For these problems you can just view the discrimiant as another algebraic expression to check, but understanding how it relates to the distance between the roots and the stationary point (y value) of the quadratic is important for explaining why the original quadratic cuts, touches or misses the x-axis. The distance between the roots intepretation drops out of the quadratic formula, so
distance = sqrt(b^2-4ac)/a
and the y value of the stationary point drops out of completing the square so
y = c - b^2/4a
and we can check whether this has the same sign as "a" (so the curve is always above or below the x-axis) nr not by multiplying by -4a, which gives the usual discriminant test.
(edited 7 months ago)

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