Quadratic equation

[Vastor]

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

 

my progression:-

[math]p^2x^2 + 3qx + 4 = 0 [/math]

 

where [math]x = a[/math] (two equal roots)

 

[math](x-a)^2 = 0[/math]

 

[math]x^2 - 2ax + a^2 = 0[/math]

compare

[math]p^2x^2 + 3qx + 4 = 0 [/math]

 

[math]a^2 = 4[/math]

 

[math]a = \pm2[/math]

 

if [math]a = 2[/math]

 

[math]-2a = 3q[/math] (compare)

 

[math]-4 = 3q[/math]

 

[math]q = \frac{-4}{3}[/math]

 

[math]\pm[/math] effect ( if a = -2)

 

[math]q = \pm\frac{4}{3}[/math]

 

but, q > 0, so no minus

 

[math]q = \frac{4}{3}[/math]

 

[math]p^2x^2 + 4x + 4 = 0[/math]

 

discriminant (two equal roots)

[math]4^2 - 4(p^2)(4) = 0[/math]

 

[math]16- 16p^2 = 0[/math]

 

[math] p = \pm 1[/math]

but, p > 0, so no minus

 

[math] p = 1[/math]

 

answer on the book

[math]p : q = 3 : 4[/math]

 

 

anyone can help?

 

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

Edited February 11, 2012 by Vastor

[Schrödinger's hat]

Didn't read much further than this:

[math]x^2 - 2ax + a^2 = 0[/math]

compare

[math]p^2 + 3qx + 4 = 0 [/math]

But you missed an [math]x^2[/math] 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 [math](x-a)^2[/math]

 

[math]x^2 + \frac{3q}{p^2}x + \frac{4}{p^2} = 0[/math]

 

So:

[math] \frac{3q}{2p^2} = a [/math] and [math] \frac{\pm 2}{p} = a[/math]

 

[math]\implies \frac{3q}{2p^2} = \frac{\pm 2}{p}\; \implies\; 3q = \pm 4p[/math]

 

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

[Vastor]

Didn't read much further than this:

 

But you missed an [math]x^2[/math] 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 [math]x^2[/math] there

 

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

 

[math]x^2 + \frac{3q}{p^2}x + \frac{4}{p^2} = 0[/math]

 

So:

[math] \frac{3q}{2p^2} = a [/math] and [math] \frac{\pm 2}{p} = a[/math]

 

[math]\implies \frac{3q}{2p^2} = \frac{\pm 2}{p}\; \implies\; 3q = \pm 4p[/math]

 

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

 

at

[math] \frac{\pm 2}{p} = a[/math]

 

why not the p has plusminus value when squareroot it?

 

EDIT : Ignore this post

Edited February 11, 2012 by Vastor

[DrRocket]

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

 

[math](x-a)^2 = 0[/math]

 

[math]C(x-a)^2 = 0[/math]

Edited February 11, 2012 by DrRocket

[Vastor]

[math]C(x-a)^2 = 0[/math]

 

simple answer that explain all, thnx

 

hope I'm doing good now:-

[math]C(x-a)^2[/math]

[math]Cx^2 - 2aCx + Ca^2 = 0 [/math]

compare

[math]p^2x^2 - 3qx +4 = 0[/math]

[math]C = p^2[/math]

 

so, [math]p^2a^2 = 4[/math]

[math]a^2 = \frac{4}{p^2}[/math]

[math]a = \frac{\pm2}{p}[/math]

 

then, [math]-2aC = -q[/math]

[math]2(\frac{\pm2}{p})(p^2) = 3q[/math]

p, q > 0 so,

[math]2(\frac{2}{p})(p^2) = 3q[/math]

[math]4p = 3q[/math]

[math]\frac{4}{3} = \frac{q}{p}[/math]

 

another problem!

 

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

 

my progression:-

[math]3x^2 + 7x - 6 = 0[/math]

 

[math]x^2 +\frac{7}{3}x - 2 = 0[/math]

compare

[math]x^2 - (a+b)x + ab = 0[/math]

 

so,

[math]a+b = -\frac{7}{3}[/math]

&

[math]ab = -2 [/math]

 

for,

[math]ab = -2 [/math]

 

[math]a = -\frac{2}{b} [/math]

 

and then, for

[math]a+b = -\frac{7}{3}[/math]

 

[math] -\frac{2}{b}+b = -\frac{7}{3}[/math]

 

[math] b = \frac{2}{b} - \frac{7}{3} [/math]

 

[math] b = \frac{6 - 7b}{3b} [/math]

 

[math]3b^2 + 7b - 6 = 0[/math]

 

[math](3b - 2)(b + 3) = 0[/math]

 

[math]b = \frac{2}{3}[/math]

and

[math]b = -3[/math]

 

because [math]b < 0[/math], so [math]b = -3[/math] only.

 

after that, for

[math]a = -\frac{2}{b} [/math]

 

[math]a = \frac{2}{3} [/math]

 

then,

[math]x^2 - (a+3 + b-2)x + (a+3)(b-2) = 0[/math]

 

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

 

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

 

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

 

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

 

[math]x^2 + \frac{4}{3}x - \frac{55}{3} = 0[/math]

 

[math]3x^2 + 4x - 55 = 0[/math]

 

and the answer is

 

[math]3x^2 + 4x - 55 = 0[/math]

 

yes, it's the same

 

but I just can't figured out how does the [math]b = \frac{2}{3} = a[/math] or it's just a coincidence?!