VCE Methods Question Thread!

[Erutepa]

Hey does any know a quick way to factorise and similar quadratics that have a coefficient that isn't 1 and cannot be simplified anyhow.
\(3x^2+5x-28\) ?

What you want to do is split the middle term into two parts whose sum is equal to 5x and whose coefficients multiply to give (3*-28=-84)

using a bit of thinking in your head, you should be able to identify that 12x and -7x add to give 5x and have coefficients that multiply to give -84

therefore you can re-write the equation as:
3x^2 + 12x - 7x - 28
 = 3x(x + 4) - 7(x + 4)
 = (3x - 7) (x - 4)

TBH I am not great at explaining maths, so please criticize my explanation (I would love to improve).
Also, I am not educated on formatting, so I apologise. I intend to learn now

[aspiringantelope]

What you want to do is split the middle term into two parts whose sum is equal to 5x and whose coefficients multiply to give (3*-28=-84)

using a bit of thinking in your head, you should be able to identify that 12x and -7x add to give 5x and have coefficients that multiply to give -84

therefore you can re-write the equation as:
3x^2 + 12x - 7x - 28
 = 3x(x + 4) - 7(x + 4)
 = (3x - 7) (x - 4)

TBH I am not great at explaining maths, so please criticize my explanation (I would love to improve).
Also, I am not educated on formatting, so I apologise. I intend to learn now

Ahh I see. Is there another and quicker way of working this out? I know this way but sometimes it's just a little slow >.<
Something I remember was like "cross something"? If anyone knows

[lzxnl]

Hey does any know a quick way to factorise and similar quadratics that have a coefficient that isn't 1 and cannot be simplified anyhow.
\(3x^2+5x-28\) ?

Quadratic formula and null factor law.

Let us solve \(3x^2 + 5x - 28 = 0\). Quadratic formula gives

So, now we know that the polynomial has these two solutions. We can write \(3x^2 + 5x - 28 = 3(x+4)\left(x - \frac{7}{3}\right)\) where the 3 at the front comes from the leading coefficient. Simplifying gives \(3x^2 + 5x - 28 = (x+4)(3x-7)\). This is if you seriously can't be bothered trying to factorise it, although you better know your square roots.

[aspiringantelope]

Quadratic formula and null factor law.

Let us solve \(3x^2 + 5x - 28 = 0\). Quadratic formula gives

So, now we know that the polynomial has these two solutions. We can write \(3x^2 + 5x - 28 = 3(x+4)\left(x - \frac{7}{3}\right)\) where the 3 at the front comes from the leading coefficient. Simplifying gives \(3x^2 + 5x - 28 = (x+4)(3x-7)\). This is if you seriously can't be bothered trying to factorise it, although you better know your square roots.

Thanks,
however there is no = 0
So I'm not sure if the Quadratic Formula also works?
What about something like \(28x^2-85x+63\)
(Just factorisation of it

[darkz]

Thanks,
however there is no = 0
So I'm not sure if the Quadratic Formula also works?
What about something like \(28x^2-85x+63\)
(Just factorisation of it

\[
\text{Let }28x^2-85x+63=0\\
\begin{aligned}
x&=\frac{85\pm \sqrt{85^2-4\times 28 \times 63}}{28*2}\\
x&=\frac{9}{7},\,\frac{7}{4}\\
\implies 28x^2-85x+63&=28(x-\frac{9}{7})(x-\frac{7}{4})\\
&=(4x-7)(7x-9)\\
\end{aligned}\\
\]

[lzxnl]

Thanks,
however there is no = 0
So I'm not sure if the Quadratic Formula also works?
What about something like \(28x^2-85x+63\)
(Just factorisation of it

I should have written 'factor theorem' there. If a polynomial P(x) satisfies P(a) = 0, then P(x) divides (x-a). So, you find all the set of points such that P(a) = 0, and these give you the linear factors of the polynomial.

[TheIllusion]

Hello AN,
I seemed to have trouble with this quadratic function.
(Apologies for possible syntax errors)

f:R+∪{0}→R, f(x)=4x^2-8x+7

Question: Express the function in the form y=a(x+b)^2+c

Thanks.

[aspiringantelope]

Hey guys! Am stuck on 9b

Got this for D if it's going to be any useful. \(D=\frac{bc}{2}\left(1-\frac{k^2}{2}\right)\)

[AlphaZero]

Hello AN,
I seemed to have trouble with this quadratic function.
(Apologies for possible syntax errors)

f:R+∪{0}→R, f(x)=4x^2-8x+7

Question: Express the function in the form y=a(x+b)^2+c

Thanks.

This question essentially asks us to complete the square:
\begin{align*}f(x)&=4\left(x^2-2x+\frac74\right)\\
&=4\left((x-1)^2-1+\frac74\right)\\
&=4(x-1)^2+3\end{align*}

Hey guys! Am stuck on 9b

Got this for D if it's going to be any useful. \(D=\frac{bc}{2}\left(1-\frac{k^2}{2}\right)\)

Assuming your result for \(D\) is correct, now just rearrange the equation to make \(k\) the subject.

[Jimmmy]

Hey guys! Am stuck on 9b

Got this for D if it's going to be any useful. \(D=\frac{bc}{2}\left(1-\frac{k^2}{2}\right)\)

Hey mate,

Not sure how to use the notation you have on the forum to explain it, but you need to use your solution for D there, and transpose it to make k the subject.

[aspiringantelope]

Hey -
So from
\(\frac{k^2}{2}=1-\frac{2D}{bc}\)
I'm not sure what to do because the answer is \(k=\sqrt{1-\frac{2D}{bc}}\) hmm
If i continue it will be like 2- .... when root 2 cannot be 1
Anyone help?

note just realised \(D=\frac{1}{2}bc\left(1-k^2\right)\) in the answers. Isn't that the same to my answer?

[AlphaZero]

Hey -
So from
\(\frac{k^2}{2}=1-\frac{2D}{bc}\)
I'm not sure what to do because the answer is \(k=\sqrt{1-\frac{2D}{bc}}\) hmm
If i continue it will be like 2- .... when root 2 cannot be 1
Anyone help?

note just realised \(D=\frac{1}{2}bc\left(1-k^2\right)\) in the answers. Isn't that the same to my answer?

They are not the same answer. You have \(k^2/2\) inside the brackets. Try part a again

[aspiringantelope]

Ok will try part A again
Here it goes
\(D=\frac{bc}{2}-\frac{bck^2}{2}\)
\(D=\frac{bc}{2}\left(1-k^2\right)\)
oooooOOOoo
I just realise I forgot to take that 2 from \(\frac{k^2}{2}\)
Damnn -_-
Probably wouldn't recover from this mistake in a real exam.
Thanks again!
And now for k
\(D=\frac{bc}{2}\left(1-k^2\right)\)
\(\frac{2D}{bc}=1-k^2\)
\(k^2=1-\frac{2D}{bc}\)
\(k=\sqrt{1-\frac{2D}{bc}}\)
Okkk!! This looks more right!!
Thanks dantraicos!!!!

I feel like my hopes of being a high scorer in methods is starting to appear again :3

[TheIllusion]

Thank you so much drantraicos.

[TheIllusion]

Hello AN,
Just had some trouble solving this cubic;
(Apologies for syntax errors)

f:[-2,4] →R
f(x)=4x^3-8x^2-16x+32

I am told to graph f(x) so I factorized the equation to 4(x-2)^2(x-2), however I cannot graph this as the (x-2)^2 touches at the same point as the (x-2) intercept.
If anyone can tell me where I went wong that would be great.
Thanks.