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P= dX^2+eX+f; d,e,f are real number, d<0 what are d, e,

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P= dX^2+eX+f; d,e,f are real number, d<0 what are d, e, [#permalink]

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22 Jul 2009, 12:52
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P= dX^2+eX+f; d,e,f are real number, d<0 what are d, e, and f?
1). When x=20, P=0
2). When x=25, P has the greatest value.

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22 Jul 2009, 13:52
vcbabu wrote:
P= dX^2+eX+f; d,e,f are real number, d<0 what are d, e, and f?
1). When x=20, P=0
2). When x=25, P has the greatest value.

should be C

stmt 1: one of the roots is = 20. insufficient

stmt 2: maxima = 25 = -e/2d.....insufficient

combine

say other root is r

sum of the roots = -e/d = 20+r
maxima = -e/2d = 25

thus, 20+r = 25*2
r = 30

You know both the roots, so now you can find d, e and f

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Manager
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22 Jul 2009, 13:53

vcbabu, sorry I don't fell safe to explain you some question without checking the answer before.

What is the OA?

PS.: Put in your other question as well please.

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23 Jul 2009, 02:01

Quote:
stmt 2: maxima = 25 = -e/2d

the question do not say that maxima is 25 !
p has the greatest value when X=25... (we do not know the greatest value)

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23 Jul 2009, 07:02

Quote:
stmt 2: maxima = 25 = -e/2d

the question do not say that maxima is 25 !
p has the greatest value when X=25... (we do not know the greatest value)

the formula -e/2d gives the value of x at maxima.

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Manager
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23 Jul 2009, 08:52
3
KUDOS
As I suspected, the fact that we know the values of two possible solutions does not determine the equation.

1)
If x = 20, when P = 0, this means that 20 is one of the solutions of the equation.
Also gives the equation 400d+20e+f=0, when substituting x and P in the equation.
And that is all
INSUFF

2) If P has the greatest value when x=25, this means that x is the x of the vertex. The x of the vertex has the formula -b/2a, considering (ax^2 + bx + c)
Thus you have -e/2d = 25.
As d<0 you have e = 50d
And that is all
INSUFF

However, if you put both stmts together, you can remember that the x of vertex is exactly between the two "zeros-solutions" of the equation.
Thus, if one solution is 20, and the x of the vertex is 25, the other solution is 30.

So you might be attempt to think: Bingo! I know that when I sum the solutions I have the opposite of e, and when I multiply them I have f. But in fact, this is only valid when the term which multiply $$x^2$$ is 1, or d=1.

In fact the right formula is
(considering ax^2 + bx + c; x1 and x2 as the solutions)
$$x1 + x2 = -b/a$$
$$x1*x2 = c/a$$

This means that since the solutions are 20 and 30, the equation is x^2 - 50x + 600
Or, because d<0, -x^2 + 50x -600
But it could be:
-2x^2 + 100x - 1200
-3x^2 + 150x - 1800
-4x^2 + 200x - 2400
.
.
.

When we read that f is a constant, this means that f is a number, but it can be -600, -1200, ...
f is a constant means that f, at the end of the solution will be a number and not a equation depending on other variables.

So, even with the two stmts together we only find a ration of d to e to f

If you liked the explanation, consider a kudo I am trying to access those GMATClub tests!!!

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23 Jul 2009, 09:13
hmmm, good for me that 2 days before the G-day I get to know I'm still lacking in few basic concepts

I always thought that if we increase the coefficients of x^2 and x along with C, they will give us a magnified graph, which will not be the same as previous one.
Thanks for the explanation coelholds. Atleast now i know one more thing.
You rightfully deserve kudos for this one

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23 Jul 2009, 09:44
P= dX^2+eX+f;
d,e,f are real number, d<0 what are d, e, and f?

1)When x=20, P=0 >> 400d+20e+f = 0 >> insufficient

2)When x=25, P has the greatest value >> maxima = 0 >> 2dx+e = 0 (take derivative as slope = 0 @ maxima) >> 50d+e=0 >> insufficient

On combining both >> since d is -ve (refer figure attached) >>

Case I is ruled out

in case II >> only single root exists ie >> both are possible 20 or 25 = -d/2e >> insufficient
in case III >> both roots are real >> so possiblity is at x=20 f(x) = 0 and at x = 25 f(x) = maximum >> insufficient

Ans E

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Bhushan S.
If you like my post....Consider it for Kudos

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24 Jul 2009, 04:48
Thanks rashminet84

Don't worry about. Unless you are a 800 test taker you will always learn something new in the previous day of the test, or even in the day. That is my opinion.

Have o good test!

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Re: quadratic DS   [#permalink] 24 Jul 2009, 04:48
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