Quick couple of question's I need some help on :)

[TommyLie]

Hello, I'm new to the forums. Just wondered if someone could help solve these for me.

Heinemann VCE zone methods 3+4 Ex1.1 Q9,10

Q9
Find solutions to 6x^3+5x^2-rx-1=0 given x=1 is a solution.

Q10
Find third solution to 3x^3+ax^2-33x-b=0 given -2 & 5 are solutions.

I know these are basic, just got confused by the fact that there is 2 & 3 unkowns involved in the equations.

Cheers
Tommy Lie

[oliverk94]

Hey, forgive me If I'm wrong, I'm not very good at Methods but I'm gonna try and give these a crack.. so if I'm wrong guys please correct me since this is learning for me.

Q9)

x=1 is a solution
Sub x=1 into the equation

6(1)^3+5(1)^2-r(1)-1=0
6(1)+5(1)-r-1=0
11-r-1=0
10-r=10
10=r
So r=10


Q10)
Here you have two unknowns and two solutions which are -2 and 5.

Sub x=-2 into equation
3(-2)^3+a(-2)^2-33(-2)-b=0
3(-8)+a(4)+66-b=0
-24+4a+66-b=0
42+4a-b=0
4a-b=-42
-b= 42-4a
b= -42+4a      [1]

Sub x=5 into equation
3(5)^3+a(5)^2-33(5)-b=0
3(125)+25a-165-b=0
375+25a-165-b=0
210+25a-b=0     [2]

Sub [1] into [2]

210+25a-(-42+4a)=0
210+25a+42-4a=0
252+21a=0
21a=-255
a= -255/21
a=-12                  [3]

Sub [3] into [1]
b=-42+4(-12)
b= -42-48
b= -90

So a=-12 and b=-90

[b^3]

Hey, forgive me If I'm wrong, I'm not very good at Methods but I'm gonna try and give these a crack.. so if I'm wrong guys please correct me since this is learning for me.

Q9)

x=1 is a solution
Sub x=1 into the equation

6(1)^3+5(1)^2-r(1)-1=0
6(1)+5(1)-r-1=0
11-r-1=0
10-r=10
10=r
So r=10


Q10)
Here you have two unknowns and two solutions which are -2 and 5.

Sub x=-2 into equation
3(-2)^3+a(-2)^2-33(-2)-b=0
3(-8)+a(4)+66-b=0
-24+4a+66-b=0
42+4a-b=0
4a-b=-42
-b= 42-4a
b= -42+4a      [1]

Sub x=5 into equation
3(5)^3+a(5)^2-33(5)-b=0
3(125)+25a-165-b=0
375+25a-165-b=0
210+25a-b=0     [2]

Sub [1] into [2]

210+25a-(-42+4a)=0
210+25a+42-4a=0
252+21a=0
21a=-255
a= -255/21
a=-12                  [3]

Sub [3] into [1]
b=-42+4(-12)
b= -42-48
b= -90

So a=-12 and b=-90

That part looks good oliverk94 but you still need to go on and find solutions to the original equations. (i.e. x=......)
i.e. for 9) you know that x=1 is a solution, so (x-1) is a factor, use long divison dividing the original by (x-1) and then factorise the quadratic you get, then from that solve for the other solutions.
for 10) x=-2 and x=5 are solutions so you know that (x+2) and (x-5) are factors, so that means that you divide the original function by (x+2)(x-5)=x²-3x-10 to get the other factor.

[oliverk94]

Ok thanks for the correction. But do you do long division before you find the unknowns or after?

[b^3]

Ok thanks for the correction. But do you do long division before you find the unknowns or after?

After, that way you are just dealing with numbers which makes it easier

[oliverk94]

Ok thanks!

[b^3]

Ok here goes nothing on the setting out........
9)

From there you know that
So use the quadratic formula for the second part



Ok that doesn't look nice, so there might me a mistake somewhere, but you get the general idea

[TommyLie]

Hey thanks for the quick replys guys! I now have managed to solve both questions. Thank you for your help

Tommy.

[oliverk94]

Ok here goes nothing on the setting out........
9)

From there you know that
So use the quadratic formula for the second part



Ok that doesn't look nice, so there might me a mistake somewhere, but you get the general idea

Yer I didn't know how to factorize 6x^2+11x+1.. do you always use the quadratic formula or can you do it with other methods like finding a factor then doing long division.. etc?

[b^3]

Ok here goes nothing on the setting out........
9)

From there you know that
So use the quadratic formula for the second part



Ok that doesn't look nice, so there might me a mistake somewhere, but you get the general idea

Yer I didn't know how to factorize 6x^2+11x+1.. do you always use the quadratic formula or can you do it with other methods like finding a factor then doing long division.. etc?

Normally the best way to do it is to sub in factors of the coefficient of the constant at the end and see what gives 0, so that you can find a factor, and do this untill you can divide through (long divison) and get a quadratic factor (i.e. if the highest power if x is 4, then you need to find 2 other factors to get to the quadratic, if its 3, then one factor). From there whichever method is the easiest to use, i.e. try to do it the easy way by regonition, i.e. looking at what times to the coefficient of the constant at the end and adds to the coeffcient of x. If that fails then I'd go for the quadratic formula.

Hope that makes sense, seem to have rambelled on a bit.

[oliverk94]

Thanks

[Planck's constant]

Yer I didn't know how to factorize 6x^2+11x+1.. do you always use the quadratic formula or can you do it with other methods like finding a factor then doing long division.. etc?

The formal way of factorising a quadratic is by the 'completing the square' method.

[b^3]

Yer I didn't know how to factorize 6x^2+11x+1.. do you always use the quadratic formula or can you do it with other methods like finding a factor then doing long division.. etc?

The formal way of factorising a quadratic is by the 'completing the square' method.

Yeh but it still depends on which methods is the most efficient, like I'm not going to factorise using completing the square (just look at what adds to 4 and mulitplies to 3, i.e. 3 and 1 so but for something like then completing the square would be a better option ().

e.g. for the above question
By the quadratic formula



By completing the square

[fred42]

All looking good except for a lost negative sign in Oliverk94's working: after 4a - b = -42, the step should be -b = -42 - 4a. The answer is then a = -8 and b = 10 giving the missing factor (3x + 1) and x = -1/3 is the last answer.

Have you considered sequential factorisation rather than long division? It's like short division. Much easier and quicker.

The answers given for Q9 are correct. I would not consider using the method of completing the square if I can use the formula. After all, the formula is derived by completing the square, so it's already done for you.

Did you know the new OS on the TI-Nspire completes the square for you?

Good luck to all those doing MM this year and hello Daniel.

[b^3]

All looking good except for a lost negative sign in Oliverk94's working: after 4a - b = -42, the step should be -b = -42 - 4a. The answer is then a = -8 and b = 10 giving the missing factor (3x + 1) and x = -1/3 is the last answer.

Yeh I just skimmed over it, and missed that negative aswell.

Have you considered sequential factorisation rather than long division? It's like short division. Much easier and quicker.

No I haven't, haven't come across it (as far as I am aware)

The answers given for Q9 are correct. I would not consider using the method of completing the square if I can use the formula. After all, the formula is derived by completing the square, so it's already done for you.

Yeh thats what I was trying to show with the different lengths/times of working out, although I probably should have stated what I was trying to show.

Did you know the new OS on the TI-Nspire completes the square for you?

Yep, refer to CAS guide in sig, but again was trying to show how much longer completing the square can be when it has to be done manually, agreed that there are better options out there than completing the square

EDIT: post number 1666