VCE Methods Question Thread!

[soNasty]

solve for x

-sqrt{1-x} -3 = -x^2-6x-8

by hand (tech free test)

[Phenomenol]

solve for x

-sqrt{1-x} -3 = -x^2-6x-8

by hand (tech free test)

Move the - 3 to the right side, then square both sides. this should get rid of the square root and leave you with a quartic, which you can factorise.

Also, after solving the quartic, make sure you can sub your x values back into the original equation (eg. getting a square root of a negative number is not good). Otherwise, reject some solutions.

[Jason12]

sketch the graphs for the following for x E(element of) [-2pi,2pi]

y=2sin3x-2

how do i find all the information need for the graph?

[soNasty]

state the value of k for which the simultaneous equations
3x-5y=8 and -6x+10y=k

have no solution and explain why

[rhinwarr]

1) 3x-5y=8
    y = (3x-8)/5

2) -6x+10y=k
    y = (6x+k)/10

For no solutions, gradients must be equal and y intercepts must be different (so there are no points of intersection).
You should know that the co-efficient of x is the gradient and the constant is the y intercept.

3/5 = 6/10      Therefore, gradients are equal.

For y intercepts to be different:
-8/5 cannot equal k/10
k cannot equal -16

So k is an element of all R\{-16}

[achre]

beaten

Let

and

Note that (2) = -2*(1), so that and .
In this situation (and you can sketch a graph to check whether or not I'm right), you can only have either infinite solutions (for k=-16, where they're the exact same line), or no solutions (for \{-16}, where they're parallel lines that never intersect).

[Rishi97]

Help pls
5) a) Sketch the graph with equation f(x) = 50/20-x where x cannot equal 20  (Done this part )
b) If g(x) = 50x/20-x :
   i ) Show that g(x) = 1000/20-x - 50

Thanks

[Phenomenol]

Help pls
5) a) Sketch the graph with equation f(x) = 50/20-x where x cannot equal 20  (Done this part )
b) If g(x) = 50x/20-x :
   i ) Show that g(x) = 1000/20-x - 50

Thanks

You want to make the numerator have a factor of (20 - x). Here's how:
50x/(20 - x) = (50x - 1000 + 1000)/(20 - x) = (-50(20 - x) + 1000)/(20 -x) = -50 + 1000/(20 - x)
as required.

[soNasty]

Consider the function


1. Find the real values of p for which the equation f(x) = p has exactly one solution.
2. Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

[alchemy]

Consider the function


1. Find the real values of p for which the equation f(x) = p has exactly one solution.
2. Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

What are the given answers?

[soNasty]

What are the given answers?

[rhinwarr]

1. Find the real values of p for which the equation f(x) = p has exactly one solution.

The solutions are where the cubic graph intersects with the horizontal line at y=p. As you can see from the graph, the value for p must be less than the local minimum turning point value or greater than the local maximum turning point value.

Spoiler

f'(x)=3x^2 - 12x + 9
For turning points, f'(x)=0
3x^2 - 12x + 9 = 0
3(x^2 - 4x + 3) = 0
3(x - 3)(x - 1) = 0
x = 3 or x = 1

f(3) = -4
f(1) = 0

Therefore, p<-4 or p>0

2. Find the real values of h for which only one of the solutions of the equation f(x + h) = 0 is positive.

f(x+h) translates the graph of f(x) h units to the left. From the graph you can see that for f(x), the two solutions are at x=1 and x=4. To make it so only one of these values are positive, you need to translate the graph enough for the x=1 solution to become negative but make sure that the solution at x=4 remains positive. This means translating the graph 1 to 4 units to the left.
Therefore, 1<=h<4

[soNasty]

thank you so much!

[~V]

Find exact value of :

[RKTR]

Find exact value of :

sin 2pi/12 =sin pi/6 =1/2