Vastor Posted February 11, 2012 Share Posted February 11, 2012 (edited) 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 Link to comment Share on other sites More sharing options...
Schrödinger's hat Posted February 11, 2012 Share Posted February 11, 2012 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. Link to comment Share on other sites More sharing options...
Vastor Posted February 11, 2012 Author Share Posted February 11, 2012 (edited) 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 Link to comment Share on other sites More sharing options...
DrRocket Posted February 11, 2012 Share Posted February 11, 2012 (edited) 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 Link to comment Share on other sites More sharing options...
Vastor Posted February 11, 2012 Author Share Posted February 11, 2012 [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?! Link to comment Share on other sites More sharing options...
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