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The equation kx^2 +4kx +3 = 0, where k is a constant, has no real roots

Prove that 0=< k < 3/4

I used the discriminant (b^2 - 4ac) to find the constant. I did find it but it was only one solution, making it an equal/= sign. Why is k less than 3/4? I'm not understanding this

Also how do you even get 0???
Reply 1
Avatar for plopapii
plopapii
2
16k^2-12k < 0 (expand disscriminant)

4k(4k-3) < 0 (factorise)

therefore,
0 < k < 3/4 (solve quadratic inequality)

if you want more explanation ask :wink:
(edited 5 years ago)
Original post by OJ Emporium
Prove that 0=< k < 3/4

I used the discriminant (b^2 - 4ac) to find the constant. I did find it but it was only one solution, making it an equal/= sign. Why is k less than 3/4? I'm not understanding this

Also how do you even get 0???


So you should have using your information 16k2−12k<016k^2 -12k<0 which simplifies and factorises to k(4k−3)<0k(4k -3)<0

Could you determine the values for which k<0? (I'd recommend sketching out the curve and looking at the inequality
Original post by OJ Emporium
Prove that 0=< k < 3/4

I used the discriminant (b^2 - 4ac) to find the constant. I did find it but it was only one solution, making it an equal/= sign. Why is k less than 3/4? I'm not understanding this

Also how do you even get 0???


So the discriminant is 16k^2 - 12k which has no real roots so it must be less than 0
16k^2 - 12k < 0
4k(4k - 3) < 0
So the roots are 0 and 3/4. If you sketch the graph of y =16k^2 - 12k, you'll find it is less than 0 when 0 < k < 3/4
So we have 0 < k < 3/4 but we want to prove 0 ≤ k < 3/4, with the 0 included, so let's try plugging in k = 0 into the first equation:
We get 0 + 0 + 3 = 0, or 3 = 0. This isn't them so k = 0 also has no real roots.
So we can combine 0 < k < 3/4 with k = 0 to get our final inequality 0 ≤ k < 3/4
Original post by 3pointonefour
So the discriminant is 16k^2 - 12k which has no real roots so it must be less than 0
16k^2 - 12k < 0
4k(4k - 3) < 0
So the roots are 0 and 3/4. If you sketch the graph of y =16k^2 - 12k, you'll find it is less than 0 when 0 < k < 3/4
So we have 0 < k < 3/4 but we want to prove 0 ≤ k < 3/4, with the 0 included, so let's try plugging in k = 0 into the first equation:
We get 0 + 0 + 3 = 0, or 3 = 0. This isn't them so k = 0 also has no real roots.
So we can combine 0 < k < 3/4 with k = 0 to get our final inequality 0 ≤ k < 3/4


Thank you for the help bud
Thanks guys
Original post by OJ Emporium
Thank you for the help bud


No problem :smile:
Original post by 3pointonefour
So the discriminant is 16k^2 - 12k which has no real roots so it must be less than 0
16k^2 - 12k < 0
4k(4k - 3) < 0
So the roots are 0 and 3/4. If you sketch the graph of y =16k^2 - 12k, you'll find it is less than 0 when 0 < k < 3/4
So we have 0 < k < 3/4 but we want to prove 0 ≤ k < 3/4, with the 0 included, so let's try plugging in k = 0 into the first equation:
We get 0 + 0 + 3 = 0, or 3 = 0. This isn't them so k = 0 also has no real roots.
So we can combine 0 < k < 3/4 with k = 0 to get our final inequality 0 ≤ k < 3/4

I've also stumbled across this question.
Why is it helpful to plug 0 back into the first equation though?


I really don't see why one would include 0 as a solution for the inequality 4k(4k-3)<0, because 0 is not less than 0, 0 is equal to zero. I think i just answered my own question.
The reason why you put k as 0 is because the inequality we could actually be solving is 4k(4k-3) ≤ 0 for which k = 0 is true.
I agree, I am pretty confused from this question I also got stuck at it.. If it has no real roots, then then b^2 -4ac must be < than 0. One of the solutions is zero and to make it work on the inequality.. it must be 0<k not =<k; if you substitute 0 into the equation you get 0<0 which is wrong.


If anyone can help, I will really appreciate it.
(edited 3 years ago)
Reply 9
Avatar for davros
davros
16
Original post by Minna Muhtaseb
I agree, I am pretty confused from this question I also got stuck at it.. If it has no real roots, then then b^2 -4ac must be < than 0. One of the solutions is zero and to make it work on the inequality.. it must be 0<k not =<k; if you substitute 0 into the equation you get 0<0 which is wrong.


If anyone can help, I will really appreciate it.

After 2 years it's hard to tell - the OP may have copied the solution down wrongly!

One possible answer is that if you put k = 0 into the original quadratic, you get 3 = 0 which obviously doesn't work! So in that sense k = 0 would also cause the equation not to have any (real) roots :smile:
Original post by davros
After 2 years it's hard to tell - the OP may have copied the solution down wrongly!

One possible answer is that if you put k = 0 into the original quadratic, you get 3 = 0 which obviously doesn't work! So in that sense k = 0 would also cause the equation not to have any (real) roots :smile:

Yeah I understand now. Thank you for helping even after 2 years 😅

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