VCE Methods Question Thread!

[qazser]

That logic is wrong.
Just because you have three solutions, doesn't mean one side is positive for values larger than the zeroes.
Example
x^2 > 1
Equality when x = -1, 1 but saying x > -1, x > 1 is wrong because x = 0 doesn't work.

Rather, you have to check in the region between each zero and see if the LHS > RHS or not in that interval. You have to check each one separately.

Yeah true, will review

[Maz]

hey guys
can someone please help me?
i know i have to do a simultaneous equation ...and ive gotton upto
8=3a+2b+c
-24=-6a+2b
y=-1a+b-c+5
i don't know if they are completely right...anyway the question is below:
The curve y=ax³+bx²+cx+5 passes through the point P (-1,4) and at the point P the first and second derivatives of the curve are 8 and -24 respectively. Find the values of the constants a,b and c
the answers are a=5, b= 3 and c=-1
thankyou 

[keltingmeith]

hey guys
can someone please help me?
i know i have to do a simultaneous equation ...and ive gotton upto
8=3a+2b+c
-24=-6a+2b
y=-1a+b-c+5
i don't know if they are completely right...anyway the question is below:
The curve y=ax³+bx²+cx+5 passes through the point P (-1,4) and at the point P the first and second derivatives of the curve are 8 and -24 respectively. Find the values of the constants a,b and c
the answers are a=5, b= 3 and c=-1
thankyou 

Erm, definitely an issue in your solutions, sorry man. The first equation doesn't work (using the given solution, 3*5+2*3+1=15+6+1=22, not , and the third still has y in it.

First, using P(-1,4), we get:
4=a(-1)^3+b(-1)^2+c(-1)+5
-1=-a+b-c <====> 1=a-b+c (same equation, just easier to write) (I'm guessing this is the third equation?)

Second, diff once, getting:
dy/dx = 3ax^2+2bx+c
Then, using the gradient at x=-1, we get:
8=3a(-1)^2+2b(-1)+c
8=3a-2b+c (I think this is what you wanted for the first equation?)

Third, diff again, getting:
d^2y/dx^2=6ax+2b
Then, using the gradient of the gradient at x=-1, we get:
-24=-6a+2b <====> 12=3a-b (yay! Second was all good)

Now, we attempt to solve the following system of equations:

1=a-b+c              (1)
8=3a-2b+c          (2)
12=3a-b              (3)

First thing I'll do is take (1) from (2), so that we lose c:

(2)-(1):
8-1=(3a-a)+(-2b+b)+(c-c)
7=2a-b                (4)

Now, we solve this by taking (3)-(4):

12-7=(3a-2a)+(-b+b)
5=a

And so we have a! Now, substitute into (3) or (4) and get:
7=2(5)-b
b=10-7=3

And now we have b! Sub into any other equation, and get:
1=5-3+c
c=1-2=-1

So, a=5, b=3, c=-1

What was my logic here?
I honestly just did random elimination things until they worked. You could also do random substitution things until they work. The reason I could get to the answer so quickly is no trick - just practice. I've been doing stuff like this for three years, after all. Don't worry - VCAA are very unlikely to ask you to solve for three variables without a CAS, and if they do, it'll be worth 2+ marks (I'd imagine), so it would be okay to spend a bit of extra time on them.

[upandgo]

hi all!  really struggling to solve these 2 problems, hopefully someone here can help out!

1) x and x+1 are two consecutive natural numbers. 1/5 of the larger exceeds 1/7 of the smaller by 1. Find the value of x.

2) The cost, $C, of electricity is determined by the number, n, units used. The rule for determining the cost is of the form C= pn+q. It is known that the cost of 200 units of electricity is $20, and of 500 units $40. Find the values of p and n.

thankyou in advance 

[qazser]

hi all!  really struggling to solve these 2 problems, hopefully someone here can help out!

1) x and x+1 are two consecutive natural numbers. 1/5 of the larger exceeds 1/7 of the smaller by 1. Find the value of x.

2) The cost, $C, of electricity is determined by the number, n, units used. The rule for determining the cost is of the form C= pn+q. It is known that the cost of 200 units of electricity is $20, and of 500 units $40. Find the values of p and n.

thankyou in advance 

Simult eqns for first question
x/5-1=x+1/7

then solve it

x=20

2. Second one is easy

To find a constant unit, take 500-200

b/c both already contain the constant addition charge (q) so the difference in them will only be units of electricity
500-200=300
40-20=20

300u=$20

therefore 1 unit of electricity is 20/300

from here should be easy

[keltingmeith]

Simult eqns for first question
x/5-1=x+1/7

then solve it

x=20

I don't think this has been quite interpreted correctly, unfortunately. Nice attempt, though!

The wording goes like so:

1) x and x+1 are two consecutive natural numbers. 1/5 of the larger exceeds 1/7 of the smaller by 1. Find the value of x.

This implies that 1/5 of the larger number and 1/7 of the smaller number should both give whole numbers. Even more, the difference between those quotients should be only 1.

21/5=4.2, 20/7=2.9

Unfortunately, neither are whole, and neither have a difference of 1.

Instead, I think the equation should be done like so:

"1/5 of the larger..."

Okay, so we start with 1/5 of the larger number (remember - the larger number is x+1, since the two numbers are consecutive):
(1/5)(x+1)=...

"...exceeds 1/7 of the smaller..."

Okay, so now we need to consider 1/7 of the smaller. There's going to be a bit more to this equation, because of that exceeds, but atm it looks like so:
(1/5)(x+1)=(1/7)x+...

"... by 1."

Alrighty, so this means that the 1/5 side will be larger by 1. That means, for the two bits to be equal, we either take 1 away from the 1/5, or add 1 to the 1/7. This gives us the following equation:
(1/5)(x+1)=(1/7)x+1

Solving simultaneously, you should get x=14. Using this, we get that:

(1/5)15=3 (this is 1/5 of the larger)
(1/7)14=2 (this is 1/7 of the smaller)

Both are whole, and the difference between them is 1. So, this solution seems to fit the bill a bit more.

Granted, this was also super tricky wording. I wouldn't expect you to have to interpret something like that in the exam. (Solve that equation? Definitely. Make that equation in the first place? Highly unlikely...)

[Maz]

Erm, definitely an issue in your solutions, sorry man. The first equation doesn't work (using the given solution, 3*5+2*3+1=15+6+1=22, not , and the third still has y in it.

First, using P(-1,4), we get:
4=a(-1)^3+b(-1)^2+c(-1)+5
-1=-a+b-c <====> 1=a-b+c (same equation, just easier to write) (I'm guessing this is the third equation?)

Second, diff once, getting:
dy/dx = 3ax^2+2bx+c
Then, using the gradient at x=-1, we get:
8=3a(-1)^2+2b(-1)+c
8=3a-2b+c (I think this is what you wanted for the first equation?)

Third, diff again, getting:
d^2y/dx^2=6ax+2b
Then, using the gradient of the gradient at x=-1, we get:
-24=-6a+2b <====> 12=3a-b (yay! Second was all good)

Now, we attempt to solve the following system of equations:

1=a-b+c              (1)
8=3a-2b+c          (2)
12=3a-b              (3)

First thing I'll do is take (1) from (2), so that we lose c:

(2)-(1):
8-1=(3a-a)+(-2b+b)+(c-c)
7=2a-b                (4)

Now, we solve this by taking (3)-(4):

12-7=(3a-2a)+(-b+b)
5=a

And so we have a! Now, substitute into (3) or (4) and get:
7=2(5)-b
b=10-7=3

And now we have b! Sub into any other equation, and get:
1=5-3+c
c=1-2=-1

So, a=5, b=3, c=-1

What was my logic here?
I honestly just did random elimination things until they worked. You could also do random substitution things until they work. The reason I could get to the answer so quickly is no trick - just practice. I've been doing stuff like this for three years, after all. Don't worry - VCAA are very unlikely to ask you to solve for three variables without a CAS, and if they do, it'll be worth 2+ marks (I'd imagine), so it would be okay to spend a bit of extra time on them.

thankyou so much

[qazser]

all i see are happy faces 

[upandgo]

I don't think this has been quite interpreted correctly, unfortunately. Nice attempt, though!

The wording goes like so:

This implies that 1/5 of the larger number and 1/7 of the smaller number should both give whole numbers. Even more, the difference between those quotients should be only 1.

21/5=4.2, 20/7=2.9

Unfortunately, neither are whole, and neither have a difference of 1.

Instead, I think the equation should be done like so:

"1/5 of the larger..."

Okay, so we start with 1/5 of the larger number (remember - the larger number is x+1, since the two numbers are consecutive):
(1/5)(x+1)=...

"...exceeds 1/7 of the smaller..."

Okay, so now we need to consider 1/7 of the smaller. There's going to be a bit more to this equation, because of that exceeds, but atm it looks like so:
(1/5)(x+1)=(1/7)x+...

"... by 1."

Alrighty, so this means that the 1/5 side will be larger by 1. That means, for the two bits to be equal, we either take 1 away from the 1/5, or add 1 to the 1/7. This gives us the following equation:
(1/5)(x+1)=(1/7)x+1

Solving simultaneously, you should get x=14. Using this, we get that:

(1/5)15=3 (this is 1/5 of the larger)
(1/7)14=2 (this is 1/7 of the smaller)

Both are whole, and the difference between them is 1. So, this solution seems to fit the bill a bit more.

Granted, this was also super tricky wording. I wouldn't expect you to have to interpret something like that in the exam. (Solve that equation? Definitely. Make that equation in the first place? Highly unlikely...)

thank you so much!! good to hear that kinda thing isnt on the exam   

[@#035;3]

solve for x:  x^2-4*a*x-5a^2=0 thanks

[keltingmeith]

solve for x:  x^2-4*a*x-5a^2=0 thanks

Oh hey, this looks like a quadratic. How can we solve quadratics?

[byCrypt]

solve for x:  x^2-4*a*x-5a^2=0 thanks

My solution:

Part 1:

Spoiler

Seeing as it's a quadratic, try factorising the expression on the left-hand side.
You should get something like (x-   )(x+   )=0

Part 2:

Spoiler

You should have got (x-5a)(x+a)=0
Using the Null Factor Law.
x-5a=0 or x+a=0
Therefore,
x=5a or x=-a

[Adequace]

Can someone point how what I did wrong for the attached, I can't seem to see it. Just need help on part b. http://imgur.com/XYozc0J

For a I got y=128/19 which was correct so for part b, my working was:

Find perpendicular gradinet of line AB, which was -1/(20/19) = -19/20, now sub point V(0,128/19) in to perpendicular line's equation which gave me y = (-19/20)x + 128/19.

However the book had a gradient of -(199/190)x

Thanks all

[Peanut Butter]

Can someone point how what I did wrong for the attached, I can't seem to see it. Just need help on part b. http://imgur.com/XYozc0J

For a I got y=128/19 which was correct so for part b, my working was:

Find perpendicular gradinet of line AB, which was -1/(20/19) = -19/20, now sub point V(0,128/19) in to perpendicular line's equation which gave me y = (-19/20)x + 128/19.

However the book had a gradient of -(199/190)x

Thanks all

The question doesn't state that the line VC is perpendicular to the line AB, therefore you can't make this assumption.

You can find the gradient of the line by using the coordinates of C (given in the question) and the coordinates of V (that you found in part a) and subbing them into the rise/run formula

Hope that helps!

[Adequace]

The question doesn't state that the line VC is perpendicular to the line AB, therefore you can't make this assumption.

You can find the gradient of the line by using the coordinates of C (given in the question) and the coordinates of V (that you found in part a) and subbing them into the rise/run formula

Hope that helps!

Ty!