DS

[prernamalhotra]

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n - m ?

(1) 4b = a2 - 4

(2) b = 0


Thank you,
Prerna

[GMATGuruNY]

If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n - m ?

(1) 4b = a^2 - 4

(2) b = 0

What is the value of n-m?
(m,0) and (n,0) are the two points where y=0 and the graph intersects the x-axis.
Thus, m and n are the x-intercepts of the graph.
Since m < n, n-m > 0.
Question rephrased: What is the positive difference between the x-intercepts?

Statement 1: 4b = a²-4
Test easy cases.

Case 1: a=0
If a=0, we get:
4b = 0²-4
b=-1.

Substituting a=0 and b=-1 into y = x² + ax + b, we get:
y = x² + 0x - 1.
y = x² - 1.
Here, y=0 when x=-1 or x=1.
Thus, the x-intercepts are -1 and 1.
Result:
Positive difference between the x-intercepts = 1 - (-1) = 2.

Case 2: a=2
If a=2, we get:
4b = 2²-4
b=0.

Substituting a=2 and b=0 into y = x² + ax + b, we get:
y = x² + 2x + 0
y = x(x+2).
Here, y=0 when x=0 or x=-2.
Thus, the x-intercepts are -2 and 0.
Result:
Positive difference between the x-intercepts = 0 - (-2) = 2.

The cases above illustrate that -- given the constraint in statement 1 -- the positive difference between the x-intercepts must be 2.
SUFFICIENT.

Statement 2: b=0
In Case 2, b=0 and a=2.
In this case, the positive difference between the x-intercepts is 2.

Case 3: b=0 and a=1
Substituting a=1 and b=0 into y = x² + ax + b, we get:
y = x² + 1x + 0
y = x(x+1).
Here, y=0 when x=0 or x=-1.
Thus, the x-intercepts are -1 and 0.
Result:
Positive difference between the x-intercepts = 0 - (-1) = 1.

Since the positive difference is not the same value in each case, INSUFFICIENT.

The correct answer is A.

[Matt@VeritasPrep]

Another approach here!

We're solving a quadratic equation of the form ax² + bx + c = 0, we know that x = (-b ± √(b² - 4ac))/2a.

Our quadratic (annoyingly) moves these letters around a bit, but plugging in appropriately, we find that x² + ax + b = 0 implies that x = (-a ± √(a² - 4b))/2.

S1 tells us that a² - 4b = 4. Plugging that into the quadratic, we have x = (-a ± √4)/2, or x = (-a ±2)/2.

Since these are the two solutions of the quadratic, the greater one is n and the smaller one is m. Thus n = (-a + 2)/2 and m = (-a - 2)/2, giving us

(n - m) = (-a + 2)/2 - (-a - 2)/2
(n - m) = (-a + a + 2 + 2)/2
(n - m) = 4/2 = 2

So S1 is SUFFICIENT!

Note that while all this stuff might seem a little foreign to the GMAT, the 2014 incarnation of the test DOES SEEM to be asking questions involving a more sophisticated understanding of quadratics -- especially the a, b, and c coefficients and the discriminant - than was previously expected. It seems to me that the question is designed to be solved in much the fashion I used above.

S2 tells us b = 0, which we can plug into our equation to get

x = (-a ± √(a² - 4*0))/2
x = (-a ± √a²)/2
x = (-a ± |a|)/2

Thus n = (-a + |a|)/2 and m = (-a - |a|)/2, giving us (n - m) = |a|. But we don't know a, so we can't solve! INSUFFICIENT

[fskilnik@GMATH]

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n - m ?

(1) 4b = a2 - 4

(2) b = 0

$$? = n - m$$
$$n > m\,\,{\rm{are}}\,\,{\rm{the}}\,\,{\rm{roots}}\,\,{\rm{of}}\,\,{\rm{the}}\,\,{\rm{equation}}\,\,{x^2} + ax + b = 0\,\,\,\left( * \right)\,\,\,\,\mathop \Rightarrow \limits^{{\rm{sum}}/{\rm{product}}} \,\,\,\,\left\{ \matrix{
\,m + n = - a \hfill \cr
\,mn = b \hfill \cr} \right.$$

$$\left( 1 \right)\,\,4b = {a^2} - 4\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\Delta = {a^2} - 4b = 4$$
$$\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,\left\{ \matrix{
\,n = {{ - a + \sqrt \Delta } \over 2} = - {a \over 2} + 1 \hfill \cr
\,m = {{ - a - \sqrt \Delta } \over 2} = - {a \over 2} - 1 \hfill \cr} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,? = n - m = 2\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,b = 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,mn = 0$$
$$\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {n,m} \right) = \left( {1,0} \right)\,\,\,\, \Rightarrow \,\,\,a = - 1\,\,\,\, \Rightarrow \,\,\,{x^2} - x = 0\,\,\,\,\left( {{\rm{viable}}} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 1 \hfill \cr
\,{\rm{Take}}\,\,\left( {n,m} \right) = \left( {2,0} \right)\,\,\,\, \Rightarrow \,\,\,a = - 2\,\,\,\, \Rightarrow \,\,\,{x^2} - 2x = 0\,\,\,\,\left( {{\rm{viable}}} \right)\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = 2 \hfill \cr} \right.$$


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.