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# Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai

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Joined: 16 Aug 2015
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Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai  [#permalink]

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21 Dec 2017, 00:48
00:00

Difficulty:

65% (hard)

Question Stats:

62% (02:20) correct 38% (02:27) wrong based on 86 sessions

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[GMAT math practice question]

Alice and Bob were asked to solve the equation $$x^2+px+q=0$$. Alice obtained the solution set $$x=1$$ and $$x=4$$, which was incorrect. She obtained the solution set of the equation $$x^2+px+s=0$$. Bob found the solution set $$x=-2$$ and $$-3$$, which was also incorrect. Bob’s answer was the solution set of the equation $$x^2+tx+q=0$$. What are the solutions of the original equation?

A. $$2,-3$$
B. $$-2,3$$
C. $$2,3$$
D. $$-2,-3$$
E. $$-1,-4$$

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"Only $79 for 1 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Retired Moderator Joined: 25 Feb 2013 Posts: 1177 Location: India GPA: 3.82 Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai [#permalink] ### Show Tags 21 Dec 2017, 03:02 1 MathRevolution wrote: [GMAT math practice question] Alice and Bob were asked to solve the equation $$x^2+px+q=0$$. Alice obtained the solution set $$x=1$$ and $$x=4$$, which was incorrect. She obtained the solution set of the equation $$x^2+px+s=0$$. Bob found the solution set $$x=-2$$ and $$-3$$, which was also incorrect. Bob’s answer was the solution set of the equation $$x^2+tx+q=0$$. What are the solutions of the original equation? A. $$2,-3$$ B. $$-2,3$$ C. $$2,3$$ D. $$-2,-3$$ E. $$-1,-4$$ to find the roots of the correct equation we need the values of $$p$$ & $$q$$ $$x^2+px+s=0$$. From this equation Sum of roots $$= \frac{-p}{1}$$ so $$-p=1+4=5=>p=-5$$ $$x^2+tx+q=0$$. from this equation Product of roots $$= \frac{q}{1}$$ So $$q=-2*-3=6$$ So correct equation is $$x^2-5x+6=0$$ Roots of the equation will be $$2$$ & $$3$$ Option C Senior SC Moderator Joined: 22 May 2016 Posts: 3574 Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai [#permalink] ### Show Tags 21 Dec 2017, 12:01 1 MathRevolution wrote: [GMAT math practice question] Alice and Bob were asked to solve the equation $$x^2+px+q=0$$. Alice obtained the solution set $$x=1$$ and $$x=4$$, which was incorrect. She obtained the solution set of the equation $$x^2+px+s=0$$. Bob found the solution set $$x=-2$$ and $$-3$$, which was also incorrect. Bob’s answer was the solution set of the equation $$x^2+tx+q=0$$. What are the solutions of the original equation? A. $$2,-3$$ B. $$-2,3$$ C. $$2,3$$ D. $$-2,-3$$ E. $$-1,-4$$ niks18 's solution is characteristically elegant. This way is more work, but very doable. Alice's and Bob's solutions were incorrect, but the equations resulting from the solutions do have a coefficient or constant that matches what we need for the original. Key variables are capitalized, thus: $$x^2+px+q=0$$ $$x^2+Px+Q=0$$ As solutions to the above, Alice incorrectly found $$x=1$$ and $$x=4$$. She obtained the solution set of the equation $$x^2+Px+s=0$$ Her solutions' resultant equation nonetheless gives us coefficient P for the original equation. (Her constant, $$s$$, is wrong. That letter must be $$q$$.) Solutions of $$x=1$$ and $$x=4$$ mean equation is $$(x-1)(x-4) = x^2 -5x + 4$$ That correlates with $$x^2+Px+s=0$$ $$P = (p = -5)$$ in original equation Bob's solution set $$x=-2$$ and $$-3$$ was also incorrect. But the solutions' resultant equation did yield constant Q. $$x^2+tx+Q=0$$ (His [t] coefficient is wrong. It is not $$p$$.) Solutions $$x=-2$$ and $$x= -3$$ yield equation $$(x + 2)(x + 3) = x^2 + 5x + 6$$ That correlates with $$x^2+tx+Q=0$$ $$Q = (q = 6)$$ in original equation Plug the correct $$p$$ and $$q$$ values into the original equation. Solve. $$x^2 - 5x + 6 = 0$$ $$(x - 2)(x - 3) = 0$$ $$x = 2$$ and $$x = 3$$ are the solutions of the original equation The answer is C (2,3) _________________ SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here. Instructions for living a life. Pay attention. Be astonished. Tell about it. -- Mary Oliver Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 8027 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai [#permalink] ### Show Tags 25 Dec 2017, 18:05 => Since $$(x-1)(x-4) = x^2 - 5x + 4 = x^2 + px +s, p = -5$$ and $$s = 4$$. Since $$(x+2)(x+3) = x^2 + 5x + 6 = x^2 + ts + q, t = 5$$ and $$q = 6$$. Thus $$x^2 + px + q = x^2 -5x + 6.$$ This factors as $$(x-2)(x-3) = 0$$, and so its solutions are $$x = 2$$ and $$x = 3$$. Therefore, the answer is C. Answer: C _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$79 for 1 month Online Course"
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Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai  [#permalink]

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13 Jan 2018, 18:59
MathRevolution wrote:
=>

Since $$(x-1)(x-4) = x^2 - 5x + 4 = x^2 + px +s, p = -5$$ and $$s = 4$$.
Since $$(x+2)(x+3) = x^2 + 5x + 6 = x^2 + ts + q, t = 5$$ and $$q = 6$$.
Thus $$x^2 + px + q = x^2 -5x + 6.$$ This factors as $$(x-2)(x-3) = 0$$, and so its solutions are $$x = 2$$ and $$x = 3$$.

Question seems wordy. It has been drafted to create confusion. Once question was clear, finding answer did not seem like 700 level, but somewhere around 600-650 level

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Alice and Bob were asked to solve the equation x^2+px+q=0. Alice obtai   [#permalink] 13 Jan 2018, 18:59
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