Polynomials Question. (1 Viewer)

[shaon0]

I haven't revised in ages so..

The expression kx^2-2x+k is negative for all values of x. Find the possible values of k.

I was using the discriminant but then don't know what to do after that.
Thanks for the help.

 

[tommykins]

回复: Polynomials Question.

Discrimintant must be < 0 and k must be < 0 (coefficient of x^2 is k, and if k is negative, it is below the x axis)

 

[shaon0]

Re: 回复: Polynomials Question.

Discrimintant must be < 0 and k must be < 0 (coefficient of x^2 is k, and if k is negative, it is below the x axis)

Thanks.
Another question on polynomials:
i) Write an expression for sin@ in terms of t=tan(0.5@)
ii) Hence solve sin@ + cot(0.5@)=2.
I am mainly having trouble with how to obtain:
sin@ in terms of t and i also don't get 't method' which may be tested (MX1 test is tomorrow)

 

[tommykins]

回复: Re: 回复: Polynomials Question.

Thanks.
Another question on polynomials:
i) Write an expression for sin@ in terms of t=tan(0.5@)
ii) Hence solve sin@ + cot(0.5@)=2.
I am mainly having trouble with how to obtain:
sin@ in terms of t and i also don't get 't method' which may be tested (MX1 test is tomorrow)

i) if t = tan(@/2) then tan@ = tan(@/2 + @/2) = 2t/1-t^2 (just expand using double angle rule).

Draw the triangle, and you will realise that the opposite hypotenuse will be 1+t^2.

sin@ = O/H = 2t/1+t^2

ii) sin@ + cot(@/2) = 2
We know that t = tan(@/2), therfore cot(@/2) = 1/t

[2t/(1-t^2)] + 1/t = 2

Multiply by 1-t^2 and then t and solve for t.

Then revert to the fact that t = tan(@/2) and solve for @.

 

[shaon0]

Re: 回复: Re: 回复: Polynomials Question.

i) if t = tan(@/2) then tan@ = tan(@/2 + @/2) = 2t/1-t^2 (just expand using double angle rule).

Draw the triangle, and you will realise that the opposite hypotenuse will be 1+t^2.

sin@ = O/H = 2t/1+t^2

ii) sin@ + cot(@/2) = 2
We know that t = tan(@/2), therfore cot(@/2) = 1/t

[2t/(1-t^2)] + 1/t = 2

Multiply by 1-t^2 and then t and solve for t.

Then revert to the fact that t = tan(@/2) and solve for @.

thanks for the help. Now i can aim for 100% in my MX1 test
EDIT: test went very badly today...looking at 50%-65% if i'm lucky