C1 Exam Question- discriminant
I was just doing a past paper, and a question went like this-
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
I think you can only find the discriminant when the equation is in the form ax^2 + bx +c
Hope this helps
Hope this helps
Original post by Ruineth
I was just doing a past paper, and a question went like this-
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
It's discriminant, not discriminate
What do you mean when you say you used the "discriminant of f(x)"? The discriminant specifically refers to an expression in terms of the standard coefficients a, b and c when the quadratic is in the form so that's the form you have to use!
Original post by Ruineth
I was just doing a past paper, and a question went like this-
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
I Don't get how to put the value of k in? I did the (x+2)^2 to get the x^2+4x+4 then did (x+2)^2+10 but the k confuses me
Original post by enaayrah
I think you can only find the discriminant when the equation is in the form ax^2 + bx +c
Hope this helps
Hope this helps
It might be useful to know that the discriminant of is .
This comes from expanding to , and then taking the discriminant as .
Original post by Ruineth
I was just doing a past paper, and a question went like this-
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
9. f(x) = x^2 + 4kx + (3 + 11k), where k is a constant.
(a) Express f(x) in the form (x + p)^2 + q, where p and q are constants to be found in terms ofk.
Given that the equation f(x) = 0 has no real roots,
(b) find the set of possible values of k.
Given that k = 1,
(c) sketch the graph of y = f(x), showing the coordinates of any point at which the graphcrosses a coordinate axis.
I did fine on a) and c)
ans to a- (x+2k)^2 -4k^2 + (3+11K)
So for b) I used the discriminate of f(x), and got the wrong answer (correct method however). They used the discriminant of the q from a).
Why did they do that?
Thanks in advance
Please give me the link to the paper, I can help.
Original post by davros
It's discriminant, not discriminate
What do you mean when you say you used the "discriminant of f(x)"? The discriminant specifically refers to an expression in terms of the standard coefficients a, b and c when the quadratic is in the form so that's the form you have to use!
What do you mean when you say you used the "discriminant of f(x)"? The discriminant specifically refers to an expression in terms of the standard coefficients a, b and c when the quadratic is in the form so that's the form you have to use!
Haha opps yeah!
I used the discriminant of the function given- x^2 + 4kx + (3 + 11k), so a=1, b=4k, c=3+11k
But they used the discriminant from the 'q' of completed square answer which was-
(x+2k)^2 -4k^2 + (3+11K),
q being -4k^2 + (3+11K),
so their a= -4k, b= 11k, c= 3
Even in the examiner's report-
Rather than using the result of part (a) to answer part (b), the vast majority used thediscriminant of the given equation.
Original post by Smaug123
It might be useful to know that the discriminant of is .
This comes from expanding to , and then taking the discriminant as .
This comes from expanding to , and then taking the discriminant as .
Ah, thanks
Original post by joshohill
Please give me the link to the paper, I can help.
https://9d9ee19f35247c2d9536ca2fef660ac833b08803.googledrive.com/host/0B1ZiqBksUHNYTzZzRjZEYVQ4TDA/11%20Gold%203%20-%20C1%20Edexcel.pdf
Question 9
Do you know that the discriminant is b2 - 4ac?
Original post by joshohill
Do you know that the discriminant is b2 - 4ac?
yeah :P
We say that when the discrim is >0 it has 2 solutions (crosses x axis twice)
We say that when the discrim is = 0 that is has 1 repeated root (NEVER crosses x axis, only touches at the repeated root) so when you factorise if you get two roots that are the same you know the curve will touch the axis at that point
We lastly can say that when the discrim is <0 there are no solutions meaning the graph never even touches or crosses the x axis.
In your question it states that f(x) has no real roots so you first right down on your page
b2 -4ac <0. that should help examiner to know where you are going and that you are using the correct method.
The discriminant requires you to work with the co-efficients of x (the bits that are written before the x).
So now plug our values of b2 -4ac into the equation using the co-efficients.
a = 1
b=4k
c=(3+11k)
carry out that calculation and you now have a quadratic that should look like this...
4k2 -11k -3
Then factorise that and you will get two values, if you have a graphical calculator it will factorise for you. If not then do it by hand; shouldnt be too hard. And you can then read off your values from your factorising part to get -1/4 k and 3.
It then needs you to put it as inequalities. This is a rule coming up so REMEMBER IT ....
if the discriminant is <0 then inequalities are ALWAYS written as 1. e.g in this question -.25<k< 3
Original post by Ruineth
Haha opps yeah!
I used the discriminant of the function given- x^2 + 4kx + (3 + 11k), so a=1, b=4k, c=3+11k
But they used the discriminant from the 'q' of completed square answer which was-
(x+2k)^2 -4k^2 + (3+11K),
q being -4k^2 + (3+11K),
so their a= -4k, b= 11k, c= 3
Even in the examiner's report-
Rather than using the result of part (a) to answer part (b), the vast majority used thediscriminant of the given equation.
I used the discriminant of the function given- x^2 + 4kx + (3 + 11k), so a=1, b=4k, c=3+11k
But they used the discriminant from the 'q' of completed square answer which was-
(x+2k)^2 -4k^2 + (3+11K),
q being -4k^2 + (3+11K),
so their a= -4k, b= 11k, c= 3
Even in the examiner's report-
Rather than using the result of part (a) to answer part (b), the vast majority used thediscriminant of the given equation.
They're using a slightly different method to get to the answer!
If you've already written f(x) in the form f(x) = (x + p)^2 + q, then solving f(x) = 0 is the same as solving (x+p)^2 + q = 0 i.e. (x+p)^2 = -q.
If q < 0 we can always take the square root of (-q) to get answers for x. So the logic they want you to follow is:
q < 0 => the equation has 2 distinct solutions
q = 0 => the equation has 1 (repeated) solution
q > 0 => the equation has no (real) solutions
Does this make things a bit clearer?
Original post by davros
They're using a slightly different method to get to the answer!
If you've already written f(x) in the form f(x) = (x + p)^2 + q, then solving f(x) = 0 is the same as solving (x+p)^2 + q = 0 i.e. (x+p)^2 = -q.
If q < 0 we can always take the square root of (-q) to get answers for x. So the logic they want you to follow is:
q < 0 => the equation has 2 distinct solutions
q = 0 => the equation has 1 (repeated) solution
q > 0 => the equation has no (real) solutions
Does this make things a bit clearer?
If you've already written f(x) in the form f(x) = (x + p)^2 + q, then solving f(x) = 0 is the same as solving (x+p)^2 + q = 0 i.e. (x+p)^2 = -q.
If q < 0 we can always take the square root of (-q) to get answers for x. So the logic they want you to follow is:
q < 0 => the equation has 2 distinct solutions
q = 0 => the equation has 1 (repeated) solution
q > 0 => the equation has no (real) solutions
Does this make things a bit clearer?
I beat you to it but I think you have a better way with words.
Original post by joshohill
I beat you to it but I think you have a better way with words.
You were quite correct in what you wrote, but I don't think the OP had a problem understanding the discriminant - I think the OP was puzzled because instead of using the discriminant from the original equation, the markscheme used the "completed square" version of the equation. I was just expanding on what I wrote earlier to make things clearer (hopefully!)
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