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March 29, 2024, 12:46:15 am

Author Topic: VCE Methods Question Thread!  (Read 4802402 times)  Share 

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lzxnl

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Re: VCE Methods Question Thread!
« Reply #17730 on: February 28, 2019, 09:32:24 pm »
+3
A family as in having the same form, a same constant (in this case)
\(y=ax^2+bx+10\)
Like having different a and b but the same constant.
Will the discriminant (in this case \(b^2+40a\) ), work with all the other parabolas in this family (same constant of 10 same form y=ax^2+bx+c, but just different number like \(y=12x^2+53x+10\) and \(y=2x^2+3x+10\) ?
edit = when we are not being given the a and b in from y=ax^2+bx+10
As mentioned above, your discriminant is of the form \(\Delta = b^2 - 40a\), and evidently if you hold a constant but change b, you get a different discriminant.

would anyone be able to help with this q please!

Given that x^3 − 2x^2 + 5 = ax(x−1)^2 + b(x−1) + c for all real numbers x, find the values of a, b and c.
This looks like partial fractions from spesh to be honest.

One way would be to expand the brackets and compare coefficients, but that's slow. Now, as the equation is true for ALL real x, you should be able to substitute in special values of x to find a, b, c. If you put in x = 1 on both sides, the first two brackets on the right hand side vanish, and you can find c. Then, put in x = 0 and the first bracket vanishes, allowing you to find b. Finally, put in any other value for x that you want to solve for a (it won't matter what new value of x you pick).
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17731 on: February 28, 2019, 10:15:02 pm »
+1
Not sure how to approach this question: "If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, what are the possible values of b?". I've tried a few thing but no luck.
 Any thoughts?

I don't really see a way other than trial and error. But, given the restrictions on the solutions, we can restrict our search down to just a few numbers with some clever observations.

First, let's solve the equation for \(x\) in terms of \(b\). By the quadratic formula, \[x=\frac{-b\pm\sqrt{b^2-60}}{2}.\] Next, both solutions for \(x\) must exist and lie in \((-10,\;10)\), so we must have \[\frac{-b-\sqrt{b^2-60}}{2}>-10\quad\text{and}\quad\frac{-b+\sqrt{b^2-60}}{2}<10\quad\text{and}\quad b^2-60>0.\] Solving those simultaneously gives: \[b\in\left(-\frac{23}{2},\;-\sqrt{60}\right)\cup\left(\sqrt{60},\;\frac{23}{2}\right).\] Now, I claim that if both solutions for \(x\) are integers, then \(b\) must also be an integer. Why? Note that we average of two integer values must be of the form \(\dfrac{k}{2}\) where \(k\in\mathbb{Z}\).  But we know that the average of the two solutions is \(\dfrac{-b}{2}\) because the graph of this quadratic is symmetrical about the line  \(x=\dfrac{-b}{2}\) (or the turning point).  So,  \(b=-k\in\mathbb{Z}\).

Noting that  \(7<\sqrt{60}<8\), we find that if we combine all the restrictions together, we have \[b\in\{-11,-10,-9,-8,\;8,\;9,\;10,\;11\}.\]Using trial and error, we obtain the final result: \[\boxed{b=\pm8}.\]
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lzxnl

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Re: VCE Methods Question Thread!
« Reply #17732 on: February 28, 2019, 11:02:37 pm »
+1
Not sure how to approach this question: "If the solutions of x^2 + bx + 15 = 0 are integers less than 10 but greater than –10, what are the possible values of b?". I've tried a few thing but no luck.
 Any thoughts?
Alternatively, you can write the equation as

If \(x_0, x_1\) are integers, and they multiply to give 15, you don't have many choices now, do you? The only four factors of 15 are 1, 3, 5, 15. Now, we want factors smaller than 10 in magnitude, so we can only take 3, 5. So, taking either \(x_0 = 3, x_1 = 5\) or \(x_0 = -3, x_0 = -5\), we get \(b = \pm 8\).
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17733 on: March 01, 2019, 10:28:37 pm »
0
How do I know if a parabola will be in the form of \(y=ax^2+c\)  or \(y=ax^2+bx\)
Thanks!
I'm trying to find the equation of a parabola from given points but i'm not sure which equation to use as i'm believing the two above both just need 2 points.

Hope someone can help me clarify.

lzxnl

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Re: VCE Methods Question Thread!
« Reply #17734 on: March 01, 2019, 11:01:06 pm »
+1
How do I know if a parabola will be in the form of \(y=ax^2+c\)  or \(y=ax^2+bx\)
Thanks!
I'm trying to find the equation of a parabola from given points but i'm not sure which equation to use as i'm believing the two above both just need 2 points.

Hope someone can help me clarify.

A parabola will always require three features. The location of a turning point counts as two, locations of coordinates count as one.

\(y = ax^2 + c\) has the vertex on the y axis. \(y = ax^2 + bx\) passes through the origin.
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17735 on: March 02, 2019, 09:45:49 am »
0
A parabola will always require three features. The location of a turning point counts as two, locations of coordinates count as one.

\(y = ax^2 + c\) has the vertex on the y axis. \(y = ax^2 + bx\) passes through the origin.

Ahhh I see! Thanks a lot!

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17736 on: March 02, 2019, 12:00:29 pm »
0
A parabola has the shape as \(y=2x^2\) but has its turning point at (1,-2). Write its equation.

leo_has_atarnotes

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Re: VCE Methods Question Thread!
« Reply #17737 on: March 02, 2019, 12:11:26 pm »
0
This question has been quite confusing for me:
Find the quotient and remainder when is divided by

The answer is:


Any thoughts?

MB_

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Re: VCE Methods Question Thread!
« Reply #17738 on: March 02, 2019, 12:46:24 pm »
0
A parabola has the shape as \(y=2x^2\) but has its turning point at (1,-2). Write its equation.
The turning point form of a parabola is \(y=a(x-h)^2+k\). You know a and (h,k) is the turning point. You should be able to go from there.
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17739 on: March 02, 2019, 01:17:51 pm »
0
The turning point form of a parabola is \(y=a(x-h)^2+k\). You know a and (h,k) is the turning point. You should be able to go from there.
But we aren't given any other points and it's in the form of \(y=2x^2)\ ?

GodNifty

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Re: VCE Methods Question Thread!
« Reply #17740 on: March 02, 2019, 01:27:56 pm »
0
But we aren't given any other points and it's in the form of \(y=2x^2)\ ?
What do you know about the dilation factor in the question? For another perspective, a normal parabola (x2) has the same form as 1(x+0)2+0
« Last Edit: March 02, 2019, 01:33:05 pm by GodNifty »

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17741 on: March 02, 2019, 01:35:54 pm »
0
What do you know about the dilation factor in the question? For another perspective, a normal parabola (x2) has the same form as 1(x+0)2+0
We are not given a from y=a(x-h)+k

GodNifty

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Re: VCE Methods Question Thread!
« Reply #17742 on: March 02, 2019, 01:41:22 pm »
0
We are not given a from y=a(x-h)+k
Yes you have.
y=2x2
2 is the dilation factor (a)
You've been given the points (1,-2) as the turning point (h is -1, -2 is k).
Now you've got a, h and k, put all these into the form of y=a(x-h)2+k

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17743 on: March 02, 2019, 02:01:53 pm »
+3
This question has been quite confusing for me:
Find the quotient and remainder when is divided by

The answer is:


Any thoughts?

Hi there. Consider this question:  "Find the quotient and remainder when 23 is divided by 5".

Well, we have  \(23=4\!\times\! 5+3\),  and so the quotient is 4 and the remainder is 3.

Method 1: Polynomial long division



The quotient is  \(1\)  (in red) and the remainder is  \(2x+2\)  (in green).

Method 2: Algebraic manipulation
\begin{align*}\frac{x^2+4}{x^2-2x+2}&=\frac{x^2+4-2x-2+2x+2}{x^2-2x+2}\\
&=\frac{x^2+4-2x-2}{x^2-2x+2}+\frac{2x+2}{x^2-2x+2}\\
&=\frac{x^2-2x+2}{x^2-2x+2}+\frac{2x+2}{x^2-2x+2}\\
&=1+\frac{2x+2}{x^2-2x+2}\\
\implies x^2+4&=1\!\times\!(x^2-2x+2)+2x+2 \end{align*} Hence, the quotient is  \(1\)  and the remainder is  \(2x+2\).
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leo_has_atarnotes

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Re: VCE Methods Question Thread!
« Reply #17744 on: March 02, 2019, 02:23:21 pm »
0
Spoiler
Hi there. Consider this question:  "Find the quotient and remainder when 23 is divided by 5".

Well, we have  \(23=4\!\times\! 5+3\),  and so the quotient is 4 and the remainder is 3.

Method 1: Polynomial long division

(Image removed from quote.)

The quotient is  \(1\)  (in red) and the remainder is  \(2x+2\)  (in green).

Method 2: Algebraic manipulation
\begin{align*}\frac{x^2+4}{x^2-2x+2}&=\frac{x^2+4-2x-2+2x+2}{x^2-2x+2}\\
&=\frac{x^2+4-2x-2}{x^2-2x+2}+\frac{2x+2}{x^2-2x+2}\\
&=\frac{x^2-2x+2}{x^2-2x+2}+\frac{2x+2}{x^2-2x+2}\\
&=1+\frac{2x+2}{x^2-2x+2}\\
\implies x^2+4&=1\!\times\!(x^2-2x+2)+2x+2 \end{align*} Hence, the quotient is  \(1\)  and the remainder is  \(2x+2\).

Thank you so much AlphaZero for your super comprehensive solution that makes a lot more sense now.
I really like how you've highlighted the latex, that's pretty clever.  :P