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800 score.com is Q> R?

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by HPengineer » Mon Dec 27, 2010 1:06 pm
Let Q and R be constants such that q>r. Let x be a real number and X^2 - 2X = (X - Q) (X - R).
What is the value of R?

A. -3
B. -2
C. 0
D. 2
E. 3

looking for some clarification.
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by anshumishra » Mon Dec 27, 2010 1:11 pm
HPengineer wrote:Let Q and R be constants such that q>r. Let x be a real number and X^2 - 2X = (X - Q) (X - R).
What is the value of R?

A. -3
B. -2
C. 0
D. 2
E. 3

looking for some clarification.
(x-q)(x-r) = x^2 - (q+r)x + qr = X^2 - 2x => qr = 0, q+r = 2
Since q>r
=> q=2, r=0
C.
Thanks
Anshu

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by HPengineer » Mon Dec 27, 2010 1:47 pm
another brilliant response!! thanks my friend i stumbled across the correct answer but my approach was not sound.

Cheers
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by Anurag@Gurome » Mon Dec 27, 2010 2:12 pm
HPengineer wrote:Let Q and R be constants such that q>r. Let x be a real number and X^2 - 2X = (X - Q) (X - R).
What is the value of R?

A. -3
B. -2
C. 0
D. 2
E. 3
Here is another method which involves factorization of given expression and comparison of the factors.

.... (x² - 2x) = (x - q)(x - r)
=> x(x - 2) = (x - q)(x - r)
=> (x - 0)(x - 2) = (x - q)(x - r)

As q > r, q = 2 and r = 0

The correct answer is C.
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by lunarpower » Mon Dec 27, 2010 2:48 pm
the poster above has grasped the intended meaning here, although the writing of the question isn't really up to snuff.
after all, the question literally says
let x be *a* real number
and, according to that wording, x could be any single real number. that's problematic because you could rig up a case to make R = just about any real number you want.
for instance, if i let x = 0 and q = 0, then the equation will always be true -- regardless of the value of r. you could even let r = -25,737,064.5523 and it would still work.

--

the wording of the question should say:
Let Q and R be constants such that q>r. If X^2 - 2X = (X - Q) (X - R) for all values of x,
what is the value of R?

if the wording of the question is changed to this one, then the correct answer is as posted by the users above. that was clearly the problem writers' intention all along, of course, but it's also good to be aware of issues in the wording.
Ron has been teaching various standardized tests for 20 years.

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