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by Rahul@gurome » Mon Dec 20, 2010 6:39 pm
We can write that y = (x+ a/2)^2 - (a^2)/4 +b.
Now (x + a/2)^2 is either 0 or positive.
So y is minimum when the above value is 0 or when (x + a/2) = 0.
This means x = -a/2 if y is minimum.
The correct answer is C.
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by Woozler » Mon Dec 20, 2010 8:49 pm
What if b = 2? Then (B) and (C) are both equally valid answers, aren't they?
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by Rahul@gurome » Mon Dec 20, 2010 9:21 pm
Woozler wrote:What if b = 2? Then (B) and (C) are both equally valid answers, aren't they?
Here "a" and "b" are constants and "x" is varying.
Obviously you have to deduce the value of x which gives the minimum value of y, irrespective of what values "a" and "b" take.
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by Night reader » Tue Dec 21, 2010 3:44 am
Good Day, below I copy/paste problem along with its official solution

If y = x^2 + ax + b, y is minimum when x is:

(a) a/b
(b) -a/b
(c) -a/2
(d) -b/2
(e) b/a

* Answer: Parabola px^2 + qx + r = 0 is minimum for x = -(q/2p). In our case y is minimum for x = -a/2.

note: as far as I remember from college math, parabola is a graphic solution of the quadratic equation.
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by anshumishra » Tue Dec 21, 2010 2:52 pm
Night reader wrote:Good Day, below I copy/paste problem along with its official solution

If y = x^2 + ax + b, y is minimum when x is:

(a) a/b
(b) -a/b
(c) -a/2
(d) -b/2
(e) b/a

* Answer: Parabola px^2 + qx + r = 0 is minimum for x = -(q/2p). In our case y is minimum for x = -a/2.

note: as far as I remember from college math, parabola is a graphic solution of the quadratic equation.
Cool !
For the Math enthusiasts (Please ignore if you don't want to use Calculus for GMAT - because it is not needed. Rahul has already solved it in a way, which any one can follow)

y = x^2+ax+b
For minimum or maximum value of y:

dy/dx = 0
=> 2x + a = 0
=> x = -a/2

To ensure whether it is minimum or max:
d/dx(dy/dx) = 2 > 0 (That means -> the value is minimum, If it were -ve, then the value would have been maximum).

Similarly you can deduce the point where parabola will have minimum value :

y = px^2 + qx + r
dy/dx = 2px+q = 0 => x = -q/2p
d/dx(dy/dx) = 2p

So, at (x = -q/2p) , parabola will have minimum value of y, if 2p > 0 and max if 2p is -ve.

Thanks
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