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# trignometry watch

1. need help in this question
i tried a couple of time and still got the same answer
so there is quadratic equation question
the thing that i got wrong is that when i solved this by factorization
i got 1 and sqroot 3=1.73.....
and when i solve by quadratic formula i got
1 and 0.73

2. (Original post by Qer)

So, what's the coefficient with ?

It's 1, so with the standard a,b,c, we have a=1

But what about the coefficient for ?

And c, the constant term = ...?
3. (Original post by ghostwalker)

So, what's the coefficient with ?

It's 1, so with the standard a,b,c, we have a=1

But what about the coefficient for ?

And c, the constant term = ...?
i solve like that
sec^2 -(1+3) tan+sq(3)=1
1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
tan^2 -sqrooot 3 tan + sqroot 3=1
tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

so from here
a=1
b= - sqroot 3
c=(sqroot 3 -1)
4. (Original post by Qer)
i solve like that
sec^2 -(1+3) tan+sq(3)=1
1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
tan^2 -sqrooot 3 tan + sqroot 3=1
tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

so from here
a=1
b= - sqroot 3
c=(sqroot 3 -1)
How did you go from your second to third line?
5. (Original post by Zacken)
How did you go from your second to third line?
1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1 ( second )
tan^2 -sqrooot 3 tan + sqroot 3=1 ( third)

expand bracket
1 + tan^2 -1 -sqroot 3 tan +sqroot 3=1 ( i cancel +1 and -1 here)
that give me that.................
6. (Original post by ghostwalker)

So, what's the coefficient with ?

It's 1, so with the standard a,b,c, we have a=1

But what about the coefficient for ?

And c, the constant term = ...?
c= sqroot 3-1?

but still not getting right answer
7. (Original post by Qer)
1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1 ( second )
tan^2 -sqrooot 3 tan + sqroot 3=1 ( third)

expand bracket
1 + tan^2 -1 -sqroot 3 tan +sqroot 3=1 ( i cancel +1 and -1 here)
that give me that.................
-(1+sq(3))tan = -tan - sq(3) tan

Not

-(1+sqrt(3))tan = -1 - sqrt(3) tan
8. (Original post by Zacken)
-(1+sq(3))tan = -tan - sq(3) tan

Not

-(1+sqrt(3))tan = -1 - sqrt(3) tan
yeah i got it after replying you

a =1
b=-1-sqroot 3
c=sqrt3 - 1

-(-1-sqroot3) +- SQRT ( -1-sqroot3)^2 -4(a)(sq root 3 -1) / 2

can you please try this for me
thanks
9. (Original post by Qer)
yeah i got it after replying you

a =1
b=-1-sqroot 3
c=sqrt3 - 1

-(-1-sqroot3) +- SQRT ( -1-sqroot3)^2 -4(a)(sq root 3 -1) / 2

can you please try this for me
thanks
Huh?

From the question and the second implication line, you have:

Why are you saying that ...?
10. (Original post by Qer)
i solve like that
sec^2 -(1+3) tan+sq(3)=1
1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
tan^2 -sqrooot 3 tan + sqroot 3=1
tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

so from here
a=1
b= - sqroot 3
c=(sqroot 3 -1)
When you change sec go tan, you can cancel 1 which appears on both sides.

This gives

a = 1

b = -(√3+1)

c = √3

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