The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Quadratics MEP1 7 (b)

7. (A) Write x2-10x+35 in the form (x-p)2+q.

I got (x-5)2+10=0

7. (B) Hence, or otherwiwe, find the maximum value of 1/(x2-10x+35)3.

My problems:
Tf is this asking? The value of x that will make the fraction largest? Surely lowest possible value of x?
Isn't it possible for negative infinity to satisfy this?
Reply 1
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
7. (A) Write x2-10x+35 in the form (x-p)2+q.

I got (x-5)2+10=0


Yep - so what this does is tells you that the quadratic has a minimum point at x=5x=5 with a value of 1010.

The reason this is so is because squares are always positive. (or zero). That is, if I take a number, whatever number I can think of and square it, I'll get a number that is ≥0\geq 0.

So - for example, let's say I pick x=−100x=-100, well then (−100)2=10000≥0(-100)^2 = 10000 \geq 0, squares are always positive.

So when you write a quadratic in the form (x−5)2+10(x-5)^2 + 10 you know that the smallest possible value of this entire expression is when the squared bracket is 0 - 'cause you can't make a squared brackets smaller than 0.

So (x−5)2≥0(x-5)^2 \geq 0 for any possible value of xx you can think of - and that means (x−5)2≥10(x-5)^2 \geq 10 for any value you can think of.

Which is basically saying that the minimum point occurs at the point (where the squared bracket becomes 0, y coordinate of that point).

The squared bracket is zero precisely when x−5=0  ⟺  x=5x-5= 0 \iff x=5, so the minimum point of this quadratic is at (5,10)(5, 10) Any other value of x will give you a y-coordinate bigger than 10.

Like, let's say x=−10x=-10 then (−10−5)2=225(-10-5)^2 = 225 so the y coordinate is 225+10≥10225 +10 \geq 10.

The graph of the quadratic looks like this:

pretty graph



So as you can see, the graph is always bigger than 10. (you might want to sketch this in your gdc in the calculator to appreciate it and understand it in the exam).


7. (B) Hence, or otherwiwe, find the maximum value of 1/(x2-10x+35)3.

My problems:
Tf is this asking? The value of x that will make the fraction largest? Surely lowest possible value of x?
Isn't it possible for negative infinity to satisfy this?


Well, you know that to make the fraction the largest, you want to make the denominator the smallest.

To make the denominator the smallest, you want to make the bit inside the bracket the smallest, because smallest^3 = smallest.

That is - we want to make x2−10x+35=(x−5)2+10x^2 - 10x + 35 = (x-5)^2 + 10 as small as possible and we showed that the smallest possible value of this is 10. If I put x = -infinity, I end up squaring that -infinity and getting +infinity^2. Which is just bad for our health. :woo:

So, yeah, the smallest value of the bracket is 10 - the smallest value of the bracket cubed is 10^3.

So the biggest value of the fraction is 1/(10^3).

Hope that makes sense, let me know if I need to explain anything anymore! :biggrin:
(edited 7 years ago)
@Zacken


Thank you :woo: I GET IT NOWWW!!
Reply 3
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
@Zacken


Thank you :woo: I GET IT NOWWW!!


Yasss! Awesome. :biggrin:

If you want to read more of my ramblings on how completing the square works (including a pretty gif) you can see here and here; but I reckon you've got it down pat. :yep:
Original post by Zacken
Yasss! Awesome. :biggrin:

If you want to read more of my ramblings on how completing the square works (including a pretty gif) you can see here and here; but I reckon you've got it down pat. :yep:

Sorry to bother you for the fifty millionth time...

How do I do this? 😓

1462539977249-523118035.jpg

At first I tried using the quadratic formula; then attempted to put it into the completed square form (to no avail)....
Original post by Zacken
Yasss! Awesome. :biggrin:

If you want to read more of my ramblings on how completing the square works (including a pretty gif) you can see here and here; but I reckon you've got it down pat. :yep:


Original post by IFoundWonderland
Sorry to bother you for the fifty millionth time...

How do I do this? 😓

1462539977249-523118035.jpg

At first I tried using the quadratic formula; then attempted to put it into the completed square form (to no avail)....


This is so cute! I love how helpful Zacken is. Wonderland, when Zacken moves to the UK for university, shall we go to Cambridge and treat him to dinner for how amazing he is?

(it's ethereal world btw)
Reply 6
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
How do I do this? 😓


If α\alpha and β\beta are roots - then you should be able to write the quadratic in the form:

(x−α)(x−β)=x2−(α+β)x+αβ=x2−kx+(k−1)(x-\alpha)(x-\beta) = x^2 -(\alpha + \beta)x + \alpha \beta = x^2 - kx + (k-1) - now you can just compare coefficients - can you take it from there?

Edit to add: whenever a question starts with "let α,β\alpha, \beta be roots..." then this particular approach should come to mind straightaway.
(edited 7 years ago)
Original post by Zacken
If α\alpha and β\beta are roots - then you should be able to write the quadratic in the form:

(x−α)(x−β)=x2−(α+β)x+αβ=x2−kx+(k−1)(x-\alpha)(x-\beta) = x^2 -(\alpha + \beta)x + \alpha \beta = x^2 - kx + (k-1) - now you can just compare coefficients - can you take it from there?

Edit to add: whenever a question starts with "let α,β\alpha, \beta be roots..." then this particular approach should come to mind straightaway.

You make it seem so simple -_-

Original post by Freudian Slit
This is so cute! I love how helpful Zacken is. Wonderland, when Zacken moves to the UK for university, shall we go to Cambridge and treat him to dinner for how amazing he is?

(it's ethereal world btw)

Haha absolutely.
https://drive.google.com/file/d/0B_iqJImnYHZ6QjlTN1pNamNjZzQ/view?usp=drivesdk
Reply 9
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
You make it seem so simple -_-


Yours is actually simpler - you were right on track with the quadratic formula.

You had:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} x = \frac{k \pm \sqrt{k^2 - 4(k-1)}}{2} = \frac{k \pm \sqrt{k^2 -4k + 4}}{2}\end{equation*}



All you need to do is notice that k2−4k+4=(k−2)2k^2 - 4k + 4 = (k-2)^2.

Note that when you expand −4(k−1)=−4k−4(−1)=4k+4≠4k−4-4(k-1) = -4k -4(-1) = 4k +4 \neq 4k-4 which you've mixed up.

Then from there on, it simplifies down to (k−2)2=k−2\sqrt{(k-2)^2} = k-2 so you get x=k±(k−2)2x = \frac{k \pm (k-2)}{2}.

Sorry - you were actually on the right track, just a sign error - ignore what I said previous about comparing coefficients, let's focus on the method you already know. :smile:
(edited 7 years ago)
Original post by Zacken
Yours is actually simpler - you were right on track with the quadratic formula.

You had:

Unparseable latex formula:

\displaystyle[br]\begin{equation*} x = \frac{k \pm \sqrt{k^2 - 4(k-1)}}{2} = \frac{k \pm \sqrt{k^2 -4k + 4}}{2}\end{equation*}



All you need to do is notice that k2−4k=4=(k−2)2k^2 - 4k = 4 = (k-2)^2 - note that when you expand −4(k−1)=−4k−4(−1)=4k+4≠4k−4-4(k-1) = -4k -4(-1) = 4k +4 \neq 4k-4.

Then from there on, it simplifies down to (k−2)2=k−2\sqrt{(k-2)^2} = k-2 so you get x=k±(k−2)2x = \frac{k \pm (k-2)}{2}.

Sorry - you were actually on the right track, just a sign error - ignore what I said previous about comparing coefficients, let's focus on the method you already know. :smile:

Hey, sorry

I don't understand why the 4 is allowed to just disappear?

If you have k^2-4k-4, this simplifies to (k-2)^2-8 ; I don't understand where the 8 vanishes to?
Reply 11
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
Hey, sorry

I don't understand why the 4 is allowed to just disappear?

If you have k^2-4k-4, this simplifies to (k-2)^2-8 ; I don't understand where the 8 vanishes to?


The thing is you don't have k^2 - 4k - 4. You have k^2 - 4k + 4.

When you expand k2−4(k−1)k^2 - 4(k-1) you get k2−4(k)−4(−1)=k2−4k+4k^2 -4(k) -4(-1) = k^2 - 4k + 4.

Just like with 5−(2−1)=5−2+1=45 - (2-1) = 5 - 2 + 1 = 4 and not 5−(2−1)=5−2−1=25 - (2-1) = 5 - 2 -1 = 2.
Original post by Zacken
The thing is you don't have k^2 - 4k - 4. You have k^2 - 4k + 4.

When you expand k2−4(k−1)k^2 - 4(k-1) you get k2−4(k)−4(−1)=k2−4k+4k^2 -4(k) -4(-1) = k^2 - 4k + 4.

Just like with 5−(2−1)=5−2+1=45 - (2-1) = 5 - 2 + 1 = 4 and not 5−(2−1)=5−2−1=25 - (2-1) = 5 - 2 -1 = 2.

OH **** MY LIFE

IT'S DIFFERENCE OF TWO SQUARES 🙄🙄🙄😂😂😂

Thanks man 👊
Reply 13
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
OH **** MY LIFE

IT'S DIFFERENCE OF TWO SQUARES 🙄🙄🙄😂😂😂

Thanks man 👊


No problem! Do you get the end answer now? You should find x=1x = 1 or x=k−1x = k - 1, so those are your roots. :biggrin:
Original post by Zacken
No problem! Do you get the end answer now? You should find x=1x = 1 or x=k−1x = k - 1, so those are your roots. :biggrin:


I absolutely do :woo: thanks for all your help :h:
Reply 15
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
I absolutely do :woo: thanks for all your help :h:


No worries. Is your exam soon? :woo:
Original post by Zacken
No worries. Is your exam soon? :woo:


Next Tuesday and Wednesday :hide:

My slip ups tend to be on small things like signs etc, so hopefully I'll iron those out and the exam will be ok.
Reply 17
Avatar for Zacken
Zacken
22
Original post by IFoundWonderland
Next Tuesday and Wednesday :hide:

My slip ups tend to be on small things like signs etc, so hopefully I'll iron those out and the exam will be ok.


I'm sure you will be. :biggrin: - Good luck! :woo:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you