anshumishra wrote:GMATGuruNY wrote:Deepthi Subbu wrote:In the xy-plane, at what two points does the graph of y = (x + a)(x + b) intersect the x-axis?
(1) a + b = -1
(2) The graph intersects the y-axis at (0, -6).
The graph will intersect the x axis when y=0.
Since y = (x+a)(x+b), y=0 when either (x+a) = 0 or (x+b) = 0.
Thus, to determine which x values will make y=0, we need to know the values of a and b.
Statement 1:
2 variables, 1 linear equation, insufficient.
Statement 2:
Substituting x=0 and y=-6 into y = (x+a)(x+b), we get:
-6 = (0+a)(0+b)
-6 = ab.
2 variables, 1 equation, insufficient.
Statements 1 and 2 together:
2 variables, 2 different equations, sufficient.
The correct answer is
C.
Not necessarily (May be we are lucky here)
If,
a+b = k1 (say 5)
a*b = k2 (say 6) --- {Also, it is not linear }
We can get two pairs of values of a and b, in general [here, (2,3) and (3,2)].
Yes, the values of a and b could be reversed. We would need to be concerned if:
-- there were a coefficient in front of either a or b in the linear equation
-- the problem asked for the value of a specifically or for the value of b specifically
For example, let's solve for the following 2 equations:
x+y = 7
xy = 12
Substituting y = 12/x into the first equation, we get:
x + 12/x = 7
x^2 + 12 = 7x
x^2 -7x + 12 = 0
(x-3)(x-4) = 0.
x = 3, 4.
Plugging these values into x+y=7, we get:
If x=3, y=4
If x=4, y=3.
Since there are no coefficients, only one combination of values will work: 3 and 4.
Now let's solve for a set of equations that includes a coefficient:
2x + y = 10.
xy = 12
Substituting y = 12/x into the first equation, we get:
2x + 12/x = 10
2x^2 + 12 = 10x
2x^2 - 10x + 12 = 0
(2x-4)(x-3) = 0.
x = 2, 3.
Plugging these values into 2x+y=10, we get:
If x=2, y=6
If x=3, y=4.
Because of the coefficient, more than one combination of values will work.
So the take-away is that given 2 equations, one of which gives the sum of 2 variables, the other of which gives the product, we'll have sufficient information if:
-- there are no coefficients
-- we need to know only what combination of values will satisfy the 2 equations
Since these 2 conditions are met in the DS question above, the 2 statements combined are sufficient, and I wouldn't waste time solving for a and b.
(Thanks for pointing out that ab = - 6 is not linear. Far from it, actually! I've amended my earlier post to ensure that no reader will be misinformed.)
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