# NSAA maths 2020 question 17

Could anyone show me how to do question 17 on the NSAA 2020 paper section A? I've tried multiple things including the quadratic formula, completing the square, and the discriminant, but I can't get the correct answer of B.

I tried screenshotting the question but it won't let me add it here
(edited 1 year ago)
For all we know, it could be something fundamental, or something silly like 3^2=6 - we can't tell.

In fact, there are a couple of ways of tackling this problem. I would like to know which way you've chosen.

(Btw, if you can't upload an image to TSR, you might want to upload to another site like imgur and post the link here)
(edited 1 year ago)
Original post by tonyiptony
For all we know, it could be something fundamental, or something silly like 3^2=6 - we can't tell.

In fact, there are a couple of ways of tackling this problem. I would like to know which way you've chosen.

(Btw, if you can't upload an image to TSR, you might want to upload to another site like imgur and post the link here)

Hi, thanks for the reply. Here is my page of workings https://imgur.com/a/AXguKTM

As you can tell I tried just proving what p might be using the discriminant being greater than 0 but I simply got a useless result from that. I tried letting x be two different possible values that differed by 6 and seeing what happened but it didn't get anywhere. I tried completing the square and using it to prove something but again I just hit a dead end and wasn't sure if I was doing anything right or not
Original post by itsskylerwhiteyo
Could anyone show me how to do question 17 on the NSAA 2020 paper section A? I've tried multiple things including the quadratic formula, completing the square, and the discriminant, but I can't get the correct answer of B.

I tried screenshotting the question but it won't let me add it here

Can you think of a method used to obtain solutions to quadratic equations?

Try applying it to the quadratic 2x^2 - px - 4 = 0 and see if you can get two expressions for x in terms of p.
(edited 1 year ago)
Original post by TypicalNerd
Can you think of a method used to obtain solutions to quadratic equations?

Try applying it to the quadratic 2x^2 - px - 4 = 0 and see if you can get two expressions for x in terms of p.

This is shown in my workings above in the imgur link. I got multiple expressions but couldn't find a way to construct a simultaneous equation
Original post by itsskylerwhiteyo
This is shown in my workings above in the imgur link. I got multiple expressions but couldn't find a way to construct a simultaneous equation

I don’t believe simultaneous equations are necessary.

Let’s start with the quadratic formula, where a = 2, b = -p and c = -4.

Can you plug these into the quadratic formula? (Note, I’m starting from the beginning because imgur is being a pain for me)
(edited 1 year ago)
Original post by TypicalNerd
I don’t believe simultaneous equations are necessary.

Let’s start with the quadratic formula, where a = 2, b = -p and c = -4.

Can you plug these into the quadratic formula? (Note, I’m starting from the beginning because imgur is being a pain for me)

Hey! Thanks for the help, I actually managed to solve it by getting my 2 x values and saying that the larger of the two was equal to the smaller of the two plus 6. Not sure if this is the best/only solution or if I overcomplicated it. TSR has now allowed me to attach my new workings

Original post by itsskylerwhiteyo
This is shown in my workings above in the imgur link. I got multiple expressions but couldn't find a way to construct a simultaneous equation

x = [ -b +/- sqrt(b^2-4ac) ] /2a
Then -b/2a is the mean of the two roots and each root differs from the mean by (one greater than and one less than)
sqrrt(b^2-4ac)/2a
So the difference between the two roots is .... and equate that to 6 and solve.for b/p.
Original post by mqb2766
x = [ -b +/- sqrt(b^2-4ac) ] /2a
Then -b/2a is the mean of the two roots and each root differs from the mean by (one greater than and one less than)
sqrrt(b^2-4ac)/2a
So the difference between the two roots is .... and equate that to 6 and solve.for b/p.

Hey I actually managed to solve the question and my finished work is just above your reply. Not sure if your explanation is the same as the way I did it or not but thanks a lot for the response and feel free to let me know if my way could be improved at all

edit Oh nevermind its being approved still thats why its not visible
(edited 1 year ago)
Original post by itsskylerwhiteyo
Hey I actually managed to solve the question and my finished work is just above your reply. Not sure if your explanation is the same as the way I did it or not but thanks a lot for the response and feel free to let me know if my way could be improved at all

edit Oh nevermind its being approved still thats why its not visible

Ahh the invisible reply ... good one :-).
Id guess they wanted you to do somehting like what I explained as the quadratic formula has a fairly simple interpretation as
x = mean +/- diff
and here they want you find b such that the overall difference is 6 or
diff = 3
So its just a slightly modified discriminant question
Original post by itsskylerwhiteyo
Hey! Thanks for the help, I actually managed to solve it by getting my 2 x values and saying that the larger of the two was equal to the smaller of the two plus 6. Not sure if this is the best/only solution or if I overcomplicated it. TSR has now allowed me to attach my new workings

Youve essentially done the same thing, but take a bit of time to get the insight, You know from the quadratic formula the roots are
x = mean +/- diff
so the difference (6) between them is
2*diff = sqrt(b^2-4ac)/a = 6
so solve from there for b/p as you do in the last 3rd of your wokring.
(edited 1 year ago)