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Can someone take a look at this.... not too bad

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Manager
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Can someone take a look at this.... not too bad [#permalink] New post 09 Oct 2006, 22:47
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A
B
C
D
E

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Can someone take a look at this.... not too bad. :lol:
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 [#permalink] New post 09 Oct 2006, 23:40
For me (D)

Y = (x+a)*(x+b)
= x^2 + (a+b)*x + a*b

Thus, Delta = (a+b)^2 - 4*a*b

We need to know whether:
o Delta < 0 (no intersect)
o Delta = 0 (a=b & 1 intersect)
o Delta > 0 (2 intersects)

Stat1:
a+b=-1

Delta = 1 - a*(-a-1)
= a^2 + a + 1
>>> Since 1^2-4*1 = -3

Thus, a^2 + a + 1 has no roots and as the coefficient of a^2 is positive,
a^2 + a + 1 > 0. Hence, Delta > 0

We have 2 roots for Y = (x+a)*(x+b) and so 2 intersects.

SUFF

Stat2:
-6 = 0^2 + (a+b)*0 + a*b
<=> a*b = -6

=> Delta = (a+b)^2 - 4*(-6)
= (a+b)^2 + 24

Hence, Delta > 0 as a sum of 2 postives numbers. We have 2 intersects.

SUFF
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 [#permalink] New post 10 Oct 2006, 01:18
where do i find the question? how to download it?
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 [#permalink] New post 10 Oct 2006, 01:19
Fig, the question asks AT WHAT TWO POINTS does the graph of the equation intersects the X axes
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 [#permalink] New post 10 Oct 2006, 09:47
BG wrote:
Fig, the question asks AT WHAT TWO POINTS does the graph of the equation intersects the X axes


He he he he ;) Yes... U are right :)... Early morning readings :)

My answer is (C) so :)
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 [#permalink] New post 10 Oct 2006, 12:48
Rayn wrote:
Can you explain Fig? Why C


(-a,0) and (-b,0) represent the 2 points that intersect the X-axis.

Actually, we have 2 unknow roots / variables. So we need 2 different equations, defining 2 relationships between a and b, to determine the values of a & b.

So, combining (1) with (2)

Equation 1 = Statment 1 : a + b = -1

Equation 2 from expliciting y= x^2 + (a+b)*x + a*b
-6= 0^2 + (a+b)*0 + a*b
<=> a*b = -6

Kind of symetrical equations.

SUFF.

a^2 + a - 6 = 0 => a=-3 or a=2 thus b=-3 or b=2.... We have the 2 points :)

Last edited by Fig on 10 Oct 2006, 14:01, edited 1 time in total.
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 [#permalink] New post 10 Oct 2006, 13:57
Given equn y=(x+a)(x+b)
If it intersects x axis it's y coordinate must be 0.

So the eqn becoms (x+a)(x+b) = 0
This is a quadratic eqn with roots -a,-b
So x=-a or x=-b.
Hence (-a,0) and (-b,0) are the two points where it intersects x-axis

Given a+b=-1 .
But from this we cannot determine a and b.
So 1 is insufficient.

Statement 2: Given that it intersects y axis at (0,-6)
Substituting in y=(x+a)(x+b) we get
ab=-6
So 2 is also insufficient.

Combining both
ie. a+b =-1 and ab=-6 we get a=-3 and b=2
S0 C must be the answer
_________________

Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

  [#permalink] 10 Oct 2006, 13:57
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