Preliminary mathematics marathon (1 Viewer)

[edmundsung]

sorry, even more confused now

so im asking the question again.
If a, b and c are consecutive positive integers show that
ax^2 + bx + c = 0 cannnot have real roots. (hint: write b and c in terms of 'a' and consider the sign of the discriminant as the sign of a quadratic.

what i have written:
Let b=a+1
c=a+2
therefore, delta=(a+1)^2 - 4a(a+2)
=-3a^2-6a+1 (corrected)

 

[hscishard]

Re: sorry, even more confused now

so im asking the question again.
If a, b and c are consecutive positive integers show that
ax^2 + bx + c = 0 cannnot have real roots. (hint: write b and c in terms of 'a' and consider the sign of the discriminant as the sign of a quadratic.

what i have written:
Let b=a+1
c=a+2
therefore, delta=(a+1)^2 - 4a(a+2)
=-3a^2-6a+1 (corrected)

Well it is a show question..

-3a^2 is < 0
Since a^2 is positive and positive x negative = negative

-6a+1 = -(6a-1). Since a > 0 then 6a-1 is > 0 (positive).
(a> 0 is given)
Negative - Positive = Even more negative
Therefore discriminant is always < 0
Therefore there is no real roots.

 

[edmundsung]

Re: sorry, even more confused now

Well it is a show question..

-3a^2 is < 0
Since a^2 is positive and positive x negative = negative

-6a+1 = -(6a-1). Since a > 0 then 6a-1 is > 0 (positive).
(a> 0 is given)
Negative - Positive = Even more negative
Therefore discriminant is always < 0
Therefore there is no real roots.

oh i get it now thanks


ps. im not smart as someone said in private msg

 

[hscishard]

Na. That was just a harder quadratic question.

 

[ohexploitable]

Re: sorry, even more confused now

oh i get it now thanks


ps. im not smart as someone said in private msg

Yeah but you're in yr 10 and doing yr 11 questions (unless you're accelerated).

 

[edmundsung]

Re: sorry, even more confused now

post deleted

 

[gurmies]

Re: sorry, even more confused now

I'm not sure if anybody solved that sin@ = 3@/4 question so here's my solution:

l = r@

r = 400/@

Other angles inside triangle are 90-@

Also note that angle at centre = 90

Using simple trig:

cos(90-@) = (400/@)/300

= 3@/4

However, cos(90-@) = sin@ ===> sin@ = 3@/4

 

[nikkifc]

Solve

for .

 

[hscishard]

Sorry exploit, I got a trig question. How can you show in b i) without using the double angle forumale?

I'm not sure if anybody solved that sin@ = 3@/4 question so here's my solution:

l = r@

r = 400/@

Other angles inside triangle are 90-@

Also note that angle at centre = 90

Using simple trig:

cos(90-@) = (400/@)/300

= 3@/4

However, cos(90-@) = sin@ ===> sin@ = 3@/4

Nice method. Omg..you gotta be a bit more clearer next time, lol.

 

[hscishard]

Yay I got it.
Draw a line that bisects 2theta.
The two triangles will be congruent, Side(radius) angle(bisected) Side(common)

x= theta
L= r x angle
400=r2x
r=200/x

Then cos(90-x) = 150/(200/x)
=3x/4
But cos(90-x) = sinx
Therefore sinx = 3x/4.

Crap. Now I forgot the double angle way.

 

[gurmies]

Nice method. Omg..you gotta be a bit more clearer next time, lol.

Sorry, I thought it was sufficiently clear.

 

[hscishard]

Sorry, I thought it was sufficiently clear.

I got lost because of the add up to 180 part and the only one @ part

 

[gurmies]

I got lost because of the add up to 180 part and the only one @ part

Oh, right. Yeah - (180-2@)/2.

Fair call.

 

[hscishard]

Wtf. I found the last one easier than the first

 

[lordinance]

PQ is a focal chord of the parabola
P is the point (6,3). Find the co-ordinates of Q.

Actually i have found Q is (6,3).

Just wondering if it is possible for both P and Q be (6,3).

thanks

 

[hscishard]

You sure?
PQ=-1
For a focal chord.

 

[ohexploitable]

You sure?
PQ=-1
For a focal chord.

uhh... pq=-1 is for parametrics, the guy was just talking about normal stuff

 

[lordinance]

oh yeah i got something wrong lol

anyway thanks

 

[hscishard]

uhh... pq=-1 is for parametrics, the guy was just talking about normal stuff

Convert it

 

[ohexploitable]

Convert it

omgd, i never thought of parametrics that way!