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