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Quick question

x/2=x-x2 Im trying to find x

Cant work it out, forgotten most of my basic algebra

Any explanation would be most appreciated!
x/2=x-x^2

x = 2(x-x^2)

x = 2x - 2x^2

0 = x - 2x^2

x - 2x^2 = 0

x(1 - 2x) = 0

Solve for x.
Reply 2
Avatar for breakeven
breakeven
OP
13
Original post by SherlockHolmes
x/2=x-x^2

x = 2(x-x^2)

x = 2x - 2x^2

0 = x - 2x^2

x - 2x^2 = 0

x(1 - 2x) = 0

Solve for x.


so x=0 or x=1/2 ?
Original post by breakeven
so x=0 or x=1/2 ?


Yes :smile:

Original post by Buttercream:)
anyone been to an interview at university of bradford for pharmacy, wondering what questions they ask :frown: really nervous interviews soon!


Wrong thread I think!
Reply 4
Avatar for breakeven
breakeven
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Thanks everyone!
Original post by SherlockHolmes
Yes :smile:



Wrong thread I think!


oh awkward -_- lmao! new to this my bad
Reply 6
Avatar for breakeven
breakeven
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13
any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?
Reply 7
Avatar for TenOfThem
TenOfThem
18
Original post by breakeven
any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?


Multiply through by cosθ then change the sin2θ
Reply 8
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breakeven
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Original post by TenOfThem
Multiply through by cosθ then change the sin2θ


How do I go about doing that?

Im at sinθx(sinθ/cosθ)-cosθ-1=0
Im trying to multiply out the brackets at the minute.
I would get sin2θ+(sinθxcosθ)-cosθ-1=0

Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)
Original post by breakeven
How do I go about doing that?

Im at sinθx(sinθ/cosθ)-cosθ-1=0
Im trying to multiply out the brackets at the minute.
I would get sin2θ+(sinθxcosθ)-cosθ-1=0

Not sure what (sinθxcosθ) is at the minute but I would then change sin2θ into (1-cosθ)


From here:
sinθx(sinθ/cosθ)-cosθ-1=0

When you multiply out the brackets, you should get sin2θ/cosθ.
Reply 10
Avatar for breakeven
breakeven
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13
Original post by SherlockHolmes
From here:
sinθx(sinθ/cosθ)-cosθ-1=0

When you multiply out the brackets, you should get sin2θ/cosθ.


So Im at sin2θ/cosθ-cosθ-1=0
Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0
Original post by breakeven
So Im at sin2θ/cosθ-cosθ-1=0
Now sure what to do from here, I need to make it the same as 2cos2θ+cosθ-1=0


See TenOfThem's post above.
Reply 12
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breakeven
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13
Original post by SherlockHolmes
See TenOfThem's post above.


I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!
Original post by breakeven
I've tried that but I'm not getting the right answer? Sorry about this, must be driving you crazy but I do appreciate it!


Post your workings so I can see where you are going wrong :smile:
Reply 14
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breakeven
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Original post by SherlockHolmes
Post your workings so I can see where you are going wrong :smile:


Man I wish I could just take a picture of my notes haha.

Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

sin2θ x cosθ -cosθ-1=0
Original post by breakeven
any insight into this question?

sinθtanθ=cosθ+1
Show this can be expressed in the form: 2cos2θ+cosθ-1=0

I took everything over to one side to get sinθtanθ -cosθ-1=0
Then i thought I should use the trig identity to get
sinθ(Sinθ/cosθ)-cosθ-1=0

Kind of stuck from here, do the sinθ's cancel out or is it sin2θ?
Also, what is sinθxcosθ ?


sin⁡θtan⁡θ=cos⁡θ+1\sin \theta \tan \theta = \cos \theta + 1

sin⁡θsin⁡θcos⁡θ=cos⁡θ+1\sin \theta \dfrac{\sin \theta}{\cos \theta} = \cos \theta + 1

sin⁡2θcos⁡θ=cos⁡θ+1\dfrac{\sin ^2 \theta}{\cos \theta} = \cos \theta + 1

As I said Multiply by cos⁡θ\cos \theta
Reply 16
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breakeven
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13
Original post by TenOfThem
sin⁡θtan⁡θ=cos⁡θ+1\sin \theta \tan \theta = \cos \theta + 1

sin⁡θsin⁡θcos⁡θ=cos⁡θ+1\sin \theta \dfrac{\sin \theta}{\cos \theta} = \cos \theta + 1

sin⁡2θcos⁡θ=cos⁡θ+1\dfrac{\sin ^2 \theta}{\cos \theta} = \cos \theta + 1

As I said Multiply by cos⁡θ\cos \theta


Oh, god I get it now. Im so sorry!
I wasn't multiplying the 1 by cosθ and i got confused, sorry and thanks so much!
Original post by breakeven
Man I wish I could just take a picture of my notes haha.

Okay so, sin2θ(sinθ/cosθ)-cosθ-1=0

sin2θ x cosθ -cosθ-1=0


sin^2θ/cosθ - when you multiply by cosθ you just get sin^2]θ.
cosθ + 1 -when you multiply that by cosθ you get cos^2θ + cosθ.
Now using the identity sin^2θ + cos^2θ = 1 you can change the sin^2θ into 1-cos^2θ (take this to the other side)
0=cos^2-1+cos^2θ+cosθ (the original information from this side)
Group like terms 2cos^2θ+cosθ-1

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