Does the equation y = (x – p)(x – q) intercept the x-axis a

[KG]

Does the equation y = (x - p)(x - q) intercept the x-axis at the point (2,0)?

(1) pq = -8

(2) -2 - p = q

I am not convinced with the explanation given in the MGMAT test.

[Guest]

OA is C.

[Stockmoose16]

(1) pq = -8

(2) -2 - p = q

I am not convinced with the explanation given in the MGMAT test.[/quote]

I think I can help you out here:

Y= (X-P)(X-Q) simplifies to:

Y= [X^2-XQ-XP]+PQ

From (1), you know pq = -8, so:

Y= [X^2 - XQ- XP] - 8

Then, you can factor the variables on the right side of the equation in the following manner:

Y= [X(X(-Q-P))] - 8

You know from (2) that -Q-P = 2

So:

Y = [X(X+2)] -8

Now, the question stem asks if the line intercepts (2,0), so plug in 2 for the X value:

Y= [2(2+2)]-8
Y=0

Therefore, YES, it does intercept at 2,0. C is correct.

[Saurav]

Does the equation y = (x - p)(x - q) intercept the x-axis at the point (2,0)?

Means that Y=0 when X = p or X = q. or otherwise (p,0) and (q,0) are the points of intersection of the curve with X axis.
(similarly if X=0, then y=pq, which means (0,pq) is the intersection with the Y-axis)

if we want to test for 2,0 then we should be able to prove that either p or q will have a value = 2

(1) pq = -8

Not Sufficient, since we cannot find one single value for p or q (following possibilities with integers, considering these are real numbers will give infinite possibilities)

(p,q)=(-1,8),(-2,4),(-4,2),(-8,1),(1,-8),(2,-4),(4,-2),(8,-1)


(2) -2 - p = q

Cannot use this to prove that p or q = 2


Now use the second equation to find out how many pairs of (p,q) follow the relation. We find that only (-4,2) will fit the relation.

Using both conditions, (-4,0) and 2,0) are the points where the curve intersects the X axis.

[RonPurewal]

perfect explanation above, except for the following:

Now use the second equation to find out how many pairs of (p,q) follow the relation. We find that only (-4,2) will fit the relation.

actually, (p, q) can be either (-4, 2) or (2, -4). i realize that you may know this, but when you write (p, q) = (_, _), the notation implies that the first number is p and the second number is q.

for any readers who don't know, these solutions can be arrived at in one of the following three ways:
(1) (formal method) use statement #2 as a substitution for q, so that pq = -8 gives p(-2 - p) = -8. solve the resulting quadratic for p, and then plug back into q = -2 - p to find q.
(2) (exhaustive listing) just try out all the possibilities for pq = -8, listed by saurav in his post, to see which ones also satisfy -2 - p = q. this may sound daunting at first, but it shouldn't seem that bad when you realize that there are only eight possibilities and the arithmetic is very easy.
(3) (inspection) just stare at the equations and see whether you can think of the numbers that satisfy them. this is a dicey method in general, and, unless your intuition for numbers and algebra is supremely fantastic, should be reserved for situations in which guessing is your only alternative.

--

by the way, the most important part of saurav's post is this buried gem:

if we want to test for 2,0 then we should be able to prove that either p or q will have a value = 2

take note: when a polynomial is written in the form (x - A)(x - B)..., the numbers A, B, ... are the x-intercepts of the polynomial's graph.

[maxschauss]

I am not sure which mistake I made, but for me (1) alone was sufficient:

First, let's look at the equation:

y = (x-p)(x-q)
y = x^2 - qx - px + pq
y = x^2 - x(q + p) + pq

The first statement says pq = -8
So plug in point (2,0) and -8:

0 = 4 - 2(q + p) - 8
0 = - 4 -2q - 2p
4 = -2q -2p
-2 = q + p (This is btw what statement (2) tells us, but I didn't know that at this point)

Now, we can plug in -2 in the above equation, so we have:

0 = 4 - 2(q + p) - 8 [--> put in -2 for q+p]
0 = 4 - 2(-2) - 8
0 = 0

Sufficient.

Statement two alone is clearly not sufficient, but why not statement one alone?

Please explain, I really can't figure out my mistake.
Thanks a lot!

[jlucero]

I am not sure which mistake I made, but for me (1) alone was sufficient:

First, let's look at the equation:

y = (x-p)(x-q)
y = x^2 - qx - px + pq
y = x^2 - x(q + p) + pq

The first statement says pq = -8
So plug in point (2,0) and -8:

0 = 4 - 2(q + p) - 8
0 = - 4 -2q - 2p
4 = -2q -2p
-2 = q + p (This is btw what statement (2) tells us, but I didn't know that at this point)

Now, we can plug in -2 in the above equation, so we have:

0 = 4 - 2(q + p) - 8 [--> put in -2 for q+p]
0 = 4 - 2(-2) - 8
0 = 0

Sufficient.

Statement two alone is clearly not sufficient, but why not statement one alone?

Please explain, I really can't figure out my mistake.
Thanks a lot!

Your math is all correct, but you need to remember that you are trying to prove that:

y = (x-p)(x-q)

There is almost like a question mark after this equation- is this actually true? So when you got it down to this step:

-2 = q + p ?

Your answer should be, only when q + p = -2. You don't know whether this is true until you add statement 2 to the mix. With one and two together, you would see that:

-2 = -2 ?

This is true, so you can answer the question with a yes. Sufficient together- (C).

[nocheivyirene]

Does the equation y = (x - p)(x - q) intercept the x-axis at the point (2,0)?

(1) pq = -8
(2) -2 - p = q

(1) pq = -8
(x - (2))(x - (-4)) where x= 2 YES!
(x -(1))(x - (-8)) where x is not 2 NO!
INSUFFICIENT!

(2) -2 -p=q means p + q = -2
x^2 + 2x + pq = 0
(x- (-1))(x- (-1)) where x is not 2 NO!
(x-(-4))(x-2) where x is 2 YES!
INSUFFICIENT!

(1) and (2) x^2 + 2x - 8 = 0 means (x+4)(x-2) =0 YES!
SUFFICIENT!

Answer: C

[jlucero]

Correct. nocheivyirene, it seems like on a lot of these problems you like choosing numbers. I'm all about that, just make sure you know when you can and can't pick numbers and how quick you need to be to be able to solve in time.

[DavidW550]

I don't see why the answer is not "D".

Using statement 1) pq=-8

Step 1) In the equation given in the question, "y=(x-p)(x-q)", expand it to read:
y=(x-p)(x-q) -> y=(x^2)-xq-xp+pq

Step 2) Plug in the values (2,0) given in the question:
0=4-2q-2p+pq

Step 3) From statement 1), pq=-8, so plug it in:
0=4-2q-2p-8 -> 4=-2q-2p -> -2=p+q

Step 4) From statement 1), q=-8/p, so:
-2=p-(8/p)

Step 5) Multiply both sides by p:
-2p=(p^2)-8 -> 0=(p^2)+2p-8 -> 0=(p+4)(p-2)
If p=-4, q=2 and if p=2, q=-4 (by plugging in p values into equation in Step 3). The equation is true when either p or q is equal to 2 and we were able to prove this by using just statement 1. SUFFICIENT

If we use statement 2), -2-p=q, we can recycle the expansion of the equation done in step 2 above: 0=4-2q-2p+pq and then. . .

Step 1) plug in statement 2), q=-2-p:
0=4-(2(-2-p))-2p+p(-2-p) -> 0=4+2p+4-2p-2p-(p^2) -> 0=8-2p-(p^2) -> 0=(4+p)(2-p)

Step 2)
If p=-4, 2 and we use the statement 2), "q=-2-p", q=2 when p=-4 and q=4 when p=2. We were once again able to prove that either p or q is equal to 2 by using just statement 2). SUFFICIENT

Why isn't the answer D?

[danr969]

the expression (x-p)(x-q) expands to x^2 -(p+q)x + pq.
We can determine the roots if we know both the product pq and the sum p+q. 1) gives the product but not the sum. 2) gives the sum but not the product. 1) and 2) together give product and sum, so C. Don't actually know if the intercept is (2,0), I just know the information is sufficient.

[RonPurewal]

dan r, do you have a question?

[RonPurewal]

DavidW550,
it appears you're trying to substitute the point (2, 0) into the equation, as though it were a known point.

the whole point of this question is to determine whether this point IS actually on the curve! so, needless to say, if you just assume it's on the curve, then you're going to conclude that ... well, it's on the curve.

[RonPurewal]

the point is that p and q are unknowns.

e.g., in statement 1, you have pq = –8.
• if one of p, q is 2 (and the other, therefore, is –4), then, yes, the point (2, 0) is on the curve.
• if p and q are any other numbers that multiply to 8, then the point (2, 0) is not on the curve.
so, statement 1 is not sufficient.
etc.

you can't use the QUESTION as though it were a FACT!

[danr969]

dan r, do you have a question?

Hi Ron,

I guess my question is whether my approach is an ok way to answer the question? Otherwise, I didn't mean to add idle chatter to the thread. Sorry for that.