#circles
88 messages · Page 1 of 1 (latest)
U have a quartic but all 4 soltutions need to be real. U can notice that since theres no cubic term if we subsitute =t^2 for u we can solve for u and then squareroot. For there to be 4 solutions u must not equal to 0 so we look for conidtions where this is true '
We first ensure the discriminant of u^2 - 6u + k is greater than 0 and then we ensure that 6 +/- sqrt(36 - 4k) is postive
6 - root 36-4k comes from the discriminant / quadratic formula
Where did u come from?
The discord
Bro i mean the letter uðŸ˜
ohh thats just choosing a variable
ill show you one sec
Thanks
oh
Thank you for the working but I’ll just firm the marks
I don’t get itðŸ˜
Just look for conditions that need to be true to have 4 roots that are real
If you square root smth less than 0 you will have a complex solution so just find and make sure that the inside of square roots are real
what paper is this?
2024
where do you find it
Bought it off the exam board
😱
how much did you buy it for
Its the mock set 4 papersðŸ˜
@tawdry pendant you could have used Discriminant as they did
I only got k>-9 when i used the discriminant
Also I think 0 in this case there is only k
Lmao
Wdym
what paper is this from
@crimson imp can we go back to this? Im trying to understand this now
So you subbed in the parametric eqns into cartesian eqn to get a quartic
Im not understanding the unique values of u bit
Yes
Youre are looking for conditions for which you will have 4 unique values of t
No repeats and all real
penaldo3142
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Yh so basically check the inside of both of the square roots and what are the requirements rhat it is not complex
And gives 2 values
Also technically its +/- for both sqrts
K>-9 is easy to spot
$$t=-sqrt{\frac{6+-\sqrt{36-4k}}{2}$$
penaldo3142
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Ok but then consider the inside off the outer sqrt that cant be negative either
So what condition has to be set there
Doesn’t k>-9 still apply for that since it’s just be 6/2
K = 9 can’t it why does it have to be less than 0
You’ve disappearedðŸ˜
Think about what happens if 6-sqrt(36-4k) is negative what happens?
Cant square root a -ve
I went to math discussion
Exactly so now what does this mean 6-sqrt(36-4k) has to be
Now keep with that idea and find another condition for k
Btw what made you use the quadratic formula?
t^4-6t^2+k=0 is what i like to call a fake quartic
Its a quartic but notice that t^4 is (t^2)^2
So you can assign t^2 to some generic variable like u
Igy sort of like a hidden quadratic
Yh
And then you can use quadratic formula and then solve using +/- sqrt
Then I just use the discriminant for the other range?
Pretty much
Alright thank you sm
Cause the first sqrt also cant be negative
Nws
Wait one more thingðŸ˜
$$\sqrt{36-4k}\textgreater0$$
$$36\textgreater4k$$
$$k\textless9$$
penaldo3142
Not minus 9ðŸ˜
It should be sqrt(36+4k)
Why plus 4k?
t^4-6t^2-k=0
I copied it down wrong earlier here 💀
Oh are we not considering the inner sqrt here?
Ohhh lol
Aight thank you