VCE Methods Question Thread!

[IndefatigableLover]

Yes I also got that discriminant I'm glad I'm not the only one who thought the question is wrong.
I also don't understand why the question asks for two rational solutions when the discriminant is a perfect square.

If the discriminant is a perfect square then you WILL get two rational solutions (if it isn't but the discriminant is still positive then you'll get an two irrational solutions). To see where I'm coming from if you used the quadratic formula and the discriminant part of it was a perfect square then it'd simplify out quite nicely and you'd get two rational solutions... maybe that's where they're coming from? I'm not 100% sure myself but if I was looking for two solutions for the equation you just gave out then I'd just quadratic formula the whole thing and hope for the best LOL >.<

I'll see what I can do later tomorrow if it's not solved by then because I'm getting tired now..

[nerdmmb]

If the discriminant is a perfect square then you WILL get two rational solutions (if it isn't but the discriminant is still positive then you'll get an two irrational solutions). To see where I'm coming from if you used the quadratic formula and the discriminant part of it was a perfect square then it'd simplify out quite nicely and you'd get two rational solutions... maybe that's where they're coming from? I'm not 100% sure myself but if I was looking for two solutions for the equation you just gave out then I'd just quadratic formula the whole thing and hope for the best LOL >.<

I'll see what I can do later tomorrow if it's not solved by then because I'm getting tired now..

That makes much more sense Thanks IndefatigableLover!

[kinslayer]

If the discriminant is a perfect square then it guarantees two rational solutions (edit: if p is rational ) but only when it isn't equal to zero, which is what I think the question is missing. If 2 = 7p then the discriminant is zero which means the equation only has one rational solution.

[Jawnle]

Evaluate using a CAS
sin(0.4)
I keep getting 0.006981 but the answer is 0.389. Something wrong with my settings?

[rhinwarr]

Evaluate using a CAS
sin(0.4)
I keep getting 0.006981 but the answer is 0.389. Something wrong with my settings?

Your calculator is in degrees but the question is in radians. I recommend changing the settings to radians permanently and when you need degrees, use the degrees symbol.

[calphanso]

Your calculator is in Gradient mode, It should be in Radians mode. I think that's the answer.

[keltingmeith]

Your calculator should be set to radians in methods, as you're more likely to work with radians than degrees. sin(0.4 degrees) = 0.006981, sin(0.4 radians) = 0.389. If you ever do work in degrees, figure out how to type in the degrees symbol on your CAS. Then, if you put that symbol into the calculator when calculating, it will treat the number given as if it were in degrees instead of radians.

[soNasty]

can someone help me derive loge(x/3)-1 please. dont know why but this is confusing me lol

***edit, what im trying to do is the entire question asking for the derivative of:

y=xloge(x/3)-x
i took out a common factor of x and had it as x(loge(x/3)-1)
and began using the product rule from there. am i on the right track though?

[lzxnl]

can someone help me derive loge(x/3)-1 please. dont know why but this is confusing me lol

d/dx (ln(x/3) - 1) = d/dx (ln x - ln 3 - 1) = 1/x
where I used a property of logs in the first step

[soNasty]

thanks lzxnl!

[EspoirTron]

can someone help me derive loge(x/3)-1 please. dont know why but this is confusing me lol

Firstly we will begin by allowing y to equal to our function; that is, y=ln(x/3)-1
Let us allow u=x/3. du/dx = 1/3
Now, y=ln(u)-1. dy/du=1/u

Hence, dy/dx = dy/du x du/dx
= 1/u x 1/3
= 1/(x/3) x 1/3 = 1/x


Thus, dy/dx = 1/x

[soNasty]

got the answer. thanks guys

[paper-back]

Is there another way to find the equation of a trigonometric graph other than visually examining it and  then guessing what the horizontal transformation may be? E.g. In the case a question wants you to find the equation in the form f(x)=sin(n(x+E))+c and they've presented you with a graph

[EspoirTron]

Is there another way to find the equation of a trigonometric graph other than visually looking at it and  then guessing what the horizontal transformation may be? E.g. In the case a question wants you to find the equation in the form f(x)=sin(n(x+E))+c and they've presented you with a graph

Yes there is. Using the points they have given you substitute them in one by one. E.g. you may have (0,1), (pi/2,0), (3pi/2,0) and (2pi,1). Substitute them in and you will get a series of simultaneous equations. You can then proceed to solve these equations. Most often you can solve them by hand but sometimes you need to use a CAS

[paper-back]

Hi Monsieur Kebab, Thanks for the quick response

I have tried substituting in values and solving using simultaneous equations, but I end up having something similar to this: E-1=E-3
So I am unable to get the horizontal transformation this way
Unless I'm doing something wrong