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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