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

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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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New post 21 Dec 2017, 00:48
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A
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C
D
E

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

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New post 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
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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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New post 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)
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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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New post 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
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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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New post 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 \(x = 3\).

Therefore, the answer is C.
Answer: C





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 &nbs [#permalink] 13 Jan 2018, 18:59
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