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Quadratic equation some question Rate Topic: -----

#1 11 February 2012 - 03:38 AM Vastor 


Baryon
The equation (px)^2 + 3qx + 4 = 0 has two equal roots. Given that p > 0 and q > 0, find p : q.

my progression:-
p^2x^2 + 3qx + 4 = 0

where x = a (two equal roots)

(x-a)^2 = 0

x^2 - 2ax + a^2 = 0
compare
p^2x^2 + 3qx + 4 = 0

a^2 = 4

a = \pm2

if a = 2

-2a = 3q (compare)

-4 = 3q

q = \frac{-4}{3}

\pm effect ( if a = -2)

q = \pm\frac{4}{3}

but, q > 0, so no minus

q = \frac{4}{3}

p^2x^2 + 4x + 4 = 0

discriminant (two equal roots)
4^2 - 4(p^2)(4) = 0

16- 16p^2 = 0

p = \pm 1
but, p > 0, so no minus

p = 1

answer on the book
p : q = 3 : 4


anyone can help? :(

edit: p^2 + 3qx + 4 = 0, sorry about the typo :/

This post has been edited by Vastor: 11 February 2012 - 04:04 AM

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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#2 11 February 2012 - 03:55 AM Schrödinger's hat 


Icon
Psychic Sexpert
Didn't read much further than this:

View PostVastor, on 11 February 2012 - 03:38 AM, said:

x^2 - 2ax + a^2 = 0
compare
p^2 + 3qx + 4 = 0

But you missed an x^2 and had a couple of odd things on later lines that looked related.

Try getting the given equation to look more like your expansion of (x-a)^2

x^2 + \frac{3q}{p^2}x + \frac{4}{p^2} = 0

So:
\frac{3q}{2p^2} = a and \frac{\pm 2}{p} = a

\implies \frac{3q}{2p^2} = \frac{\pm 2}{p}\; \implies\; 3q = \pm 4p

The \pm must be + or we don't have both p and q > 0.
I don't believe in free will, but I choose to pretend it exists. If I'm helpful press the green button--->
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#3 11 February 2012 - 04:15 AM Vastor 


Baryon

View PostSchrödinger, on 11 February 2012 - 03:55 AM, said:

Didn't read much further than this:

But you missed an x^2 and had a couple of odd things on later lines that looked related.


not really, just typo transferring from the paper that I used to calculate, and not missing any x^2 there

View PostSchrödinger, on 11 February 2012 - 03:55 AM, said:

Try getting the given equation to look more like your expansion of (x-a)^2

x^2 + \frac{3q}{p^2}x + \frac{4}{p^2} = 0

So:
\frac{3q}{2p^2} = a and \frac{\pm 2}{p} = a

\implies \frac{3q}{2p^2} = \frac{\pm 2}{p}\; \implies\; 3q = \pm 4p

The \pm must be + or we don't have both p and q > 0.


at
\frac{\pm 2}{p} = a

why not the p has plusminus value when squareroot it?

EDIT : Ignore this post :D

This post has been edited by Vastor: 11 February 2012 - 06:06 AM

Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
0

#4 11 February 2012 - 05:11 AM DrRocket 


Primate

View PostVastor, on 11 February 2012 - 03:38 AM, said:

The equation (px)^2 + 3qx + 4 = 0 has two equal roots. Given that p > 0 and q > 0, find p : q.

(x-a)^2 = 0


C(x-a)^2 = 0

This post has been edited by DrRocket: 11 February 2012 - 05:11 AM


You can know the name of a bird in all the languages of the world, but when you're finished, you'll know absolutely nothing whatever about the bird... -- Richard P. Feynman
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#5 11 February 2012 - 08:50 AM Vastor 


Baryon

View PostDrRocket, on 11 February 2012 - 05:11 AM, said:

C(x-a)^2 = 0


simple answer that explain all, thnx :lol:

hope I'm doing good now:-
C(x-a)^2
Cx^2 - 2aCx + Ca^2 = 0
compare
p^2x^2 - 3qx +4 = 0
C = p^2

so, p^2a^2 = 4
a^2 = \frac{4}{p^2}
a = \frac{\pm2}{p}

then, -2aC = -q
2(\frac{\pm2}{p})(p^2) = 3q
p, q > 0 so,
2(\frac{2}{p})(p^2) = 3q
4p = 3q
\frac{4}{3} = \frac{q}{p}

another problem!<_<

given that a and b are the roots of the equation 3x^2 + 7x - 6 = 0 where a > 0 and b < 0. Form a quadratic equation which has the roots a + 3 and b-2.

my progression:-
3x^2 + 7x - 6 = 0

x^2 +\frac{7}{3}x - 2 = 0
compare
x^2 - (a+b)x + ab = 0

so,
a+b = -\frac{7}{3}
&
ab = -2

for,
ab = -2

a = -\frac{2}{b}

and then, for
a+b = -\frac{7}{3}

-\frac{2}{b}+b = -\frac{7}{3}

b = \frac{2}{b} - \frac{7}{3}

b = \frac{6 - 7b}{3b}

3b^2 + 7b - 6 = 0

(3b - 2)(b + 3) = 0

b = \frac{2}{3}
and
b = -3

because b < 0, so b = -3 only.

after that, for
a = -\frac{2}{b}

a = \frac{2}{3}

then,
x^2 - (a+3 + b-2)x + (a+3)(b-2) = 0

x^2 -(-\frac{7}{3} + 3-2)x + (ab+3b - 2a-6) = 0

x^2 + \frac{4}{3}x +(-8 + 3b - 2a) = 0

x^2 + \frac{4}{3}x +(-8 + 3(-3) - 2(\frac{2}{3})) = 0

x^2 + \frac{4}{3}x +(-17 - (\frac{4}{3})) = 0

x^2 + \frac{4}{3}x - \frac{55}{3} = 0

3x^2 + 4x - 55 = 0

and the answer is

3x^2 + 4x - 55 = 0

yes, it's the same :P

but I just can't figured out how does the b = \frac{2}{3} = a or it's just a coincidence?! :huh:
Some sort of good tutor :}

the one who don't bother to accept is ignorant, the one who don't bother to deny is fools.
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