Mathematics Question Thread

[Mathew587]

Hey! For the first one to have real and rational roots, we just need to prove that the discriminant (\\Delta\) is greater than or equal to zero, and that it is a perfect square.



Clearly, in that factorised form, this will never be a negative number (squared numbers are always positive). So, since the discriminant is never negative, there are always real roots. Since it is a perfect square, they are always rational

For that second one, consider the general quadratic \(y=ax^2+bx+c\), and substitute the three pairs of coordinates in to get three equations in terms of \(a\), \(b\) and \(c\) - Solve those simultaneously if you can (otherwise post where you get stuck and we can guide you the rest of the way!)

I followed what you told me and got:
18-4a+2b-c=0 for (-2,18) and -2-9a-3b-c=0 for (3,-2)
I then equated c's and got 20=-5a-5b - (1)
I then got 0=a+b+c for (1,0), found a and subbed that into the (1) equation and got c=4 which is wrong.
Can you please help me

[RuiAce]

I followed what you told me and got:
18-4a+2b-c=0 for (-2,18) and -2-9a-3b-c=0 for (3,-2)
I then equated c's and got 20=-5a-5b - (1)
I then got 0=a+b+c for (1,0), found a and subbed that into the (1) equation and got c=4 which is wrong.
Can you please help me

[Mathew587]

Yup i did that and continued by subbing in (1,0) into the quadratic as previously stated but it didn't work out. Any working out for that?

[RuiAce]

Yuck. Sorry, seems I had made a transcription error in my working out.

Which gets what you got. And when I checked on WolframAlpha c=4 is correct.

(From here, just sub c=4 into any two of the three other equations, and solve them to find a and b)

[Mathew587]

Yuck. Sorry, seems I had made a transcription error in my working out.

Which gets what you got. And when I checked on WolframAlpha c=4 is correct.

(From here, just sub c=4 into any two of the three other equations, and solve them to find a and b)

Oh ok. My confusion came from the fact that the answers had a different set of solutions with a=4, b=-3, c=7. Does that still make sense with the equation?

[RuiAce]

Oh ok. My confusion came from the fact that the answers had a different set of solutions with a=4, b=-3, c=7. Does that still make sense with the equation?

I checked the textbook. You're looking at the answers to Q3. Not Q2.

You get \(a=1, b=-5, c=4\) which when you sub it all back in to \(ax^2+bx+c\) you have \(x^2-5x+4\)

[bananna]

hi!

With this q attached, I understand everything until the graph
I'm not completely sure how to draw it
any help would be appreciated ..thank you!

[RuiAce]

hi!

With this q attached, I understand everything until the graph
I'm not completely sure how to draw it
any help would be appreciated ..thank you!

Can you please list the answers to the previous parts? Because all of them (except maybe a) and f)) will be necessary to draw the final sketch

[bananna]

Can you please list the answers to the previous parts? Because all of them (except maybe a) and f)) will be necessary to draw the final sketch

yes! sorry about that

a) y'=e^x(1+x)
    y''= e^x(2+x)

b) minimum Stat.P at (-1,-e^-1)

c) p.o.i at (-2,-2e^-2)

d) curve passes thru origin

e) as x --> infinity
        y--> infinity

f) y --> 0
   y ---> 0

thanks

[Shadowxo]

yes! sorry about that

a) y'=e^x(1+x)
    y''= e^x(2+x)

b) minimum Stat.P at (-1,-e^-1)

c) p.o.i at (-2,-2e^-2)

d) curve passes thru origin

e) as x --> infinity
        y--> infinity

f) y --> 0
   y ---> 0

thanks

Can't do a full solution with a graph, but we know
There are points at (-1,-e^-1), (-2,-2e^-2), and (0,0). We know as x becomes more negative, y approaches 0 from below, so y=0 is an asymptote. We know there's a point of inflection at (-2,-2e^-2), and a minimum TP at (-1,-e^-1), then it goes up, passes through the origin and as x->infinity y-> infinity. Also, as e>2, we know -2e^-2< -e^-1 (closer to y=0). Now we have all the information to graph it. Starts close to y=0 as x is very negative, then as x gets larger the y value goes down slightly, until (-2,-2e^-2) where there's a small point of inflection, then continues down, then passes through (-1,-e^-1) and starts going up, passes through (0,0) and continues upwards.

[RuiAce]

yes! sorry about that

a) y'=e^x(1+x)
    y''= e^x(2+x)

b) minimum Stat.P at (-1,-e^-1)

c) p.o.i at (-2,-2e^-2)

d) curve passes thru origin

e) as x --> infinity
        y--> infinity

f) y --> 0
   y ---> 0

thanks

Here is a simple GeoGebra simulation of the graph, including many important details you have discovered. Note clearly the intercept, turning point and inflexion point provided, as well as the fact that the remainder of the question builds up to the asymptote y=0 for really large negative values of x.

[Mathew587]

I checked the textbook. You're looking at the answers to Q3. Not Q2.

You get \(a=1, b=-5, c=4\) which when you sub it all back in to \(ax^2+bx+c\) you have \(x^2-5x+4\)

Oh my god rui sorry about that    Ive only had 5hrs of sleep and my heads on fire with the upcoming half yearlies. Thank you anyways
Mathew

[katnisschung]

18) two identical cubes (similar to dice) each having faces 0-5 are rolled.
A score for the roll is determined as the product of the two numbers on the two uppermost faces.

b) if the cubes are rolled twice and the scores for each roll are added, what is the probability of a
combined score of at least 41

my answer= 5/1296

actual answer= 7/1296

I don't understand why counted the scores of the following (2 are repeated?)

scores of (16,25), (25,16), (25,25), (20, 25), (25,20)
(20, 25), (25,20)--> why are these two repeated?? is this an error?

[Arisa_90]

Are there particular rules for differentiating trig and exponential?
I am not sure what to do when asked to differentiate tan

[RuiAce]

Are there particular rules for differentiating trig and exponential?
I am not sure what to do when asked to differentiate tan

18) two identical cubes (similar to dice) each having faces 0-5 are rolled.
A score for the roll is determined as the product of the two numbers on the two uppermost faces.

b) if the cubes are rolled twice and the scores for each roll are added, what is the probability of a
combined score of at least 41

my answer= 5/1296

actual answer= 7/1296

I don't understand why counted the scores of the following (2 are repeated?)

scores of (16,25), (25,16), (25,25), (20, 25), (25,20)
(20, 25), (25,20)--> why are these two repeated?? is this an error?

We have to include the repetition.

Getting 4 and 5 on the first try and getting 5 and 5 on the second try, is DIFFERENT to getting 5 and 5 on the first try and 4/5 on the second here. They're two distinct ways of achieving the same final outcome. EDIT: I'm tired and I misread. Reanalysing.

I believe the repetition occurs because in just the first try, there's TWO ways of getting a 4 and a 5. The first die can roll the 4, but so can the second.

Whereas there's only one way of getting two 4's