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

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Math Revolution GMAT Instructor
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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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20 Dec 2017, 23:48
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Question Stats:

55% (11:59) correct 45% (01:31) wrong based on 69 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. $$-1,-4$$
[Reveal] Spoiler: OA

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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, 02:02
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
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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, 11:01
1
KUDOS
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

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

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25 Dec 2017, 17:05
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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 $$x = 3$$.

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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, 17:59
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, 17:59
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