X-axis

[gmatblood]

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

(1) pq = -8

(2) -2 - p = q

[rijul007]

y = (x-p)(x-q)
For x-intercept to be (2,0)

0 = (2-p)(2-q)

So we need to check whether either of p or q is 2...

Statement 1
pq = -8

value of p and q can b anything
-1*8
-8*1
2*-4
...

Insufficient


Statement 2
-2 - p = q

We are still not sure of values of p and q



Combining Statement 1 and 2
pq = -8
-2 -p = q
p = -4
q = 2
therefore x-intercept of the equation is (2,0)


Option C

[gmatblood]

Hi,

Could you please explain this part,

For x-intercept to be (2,0)

0 = (2-p)(2-q)

So we need to check whether either of p or q is 2...

thanks.

[rijul007]

Hi,

Could you please explain this part,

For x-intercept to be (2,0)

0 = (2-p)(2-q)

So we need to check whether either of p or q is 2...

thanks.

The equation is
y = (x-p)(x-q)

What do we do to find the x-intercept???
x-intercept is supposed to lie on the x-axis, which is y = 0.

so we substitute y=0 in the equation,
0 = (x-p)(x-q)
This equation tells us that value of x should either be equal to p or q......

Acc to the ques., x-intercept = (2,0)
here x = 2

This means either p or q has to be 2.

Now you need to check which statement tells that...

[HSPA]

I have y = x^2 - (p+q)x + pq

From option 1: I have pq
From option 2: I have p+q

Thus I can solve the quadratic equation x^2+2x-8 = (x+4)(x-2)

Thus when x =2 y=0 hence proved..

[zeusunlimited]

my 0.02$
for P(2,0),
y=0=(x-p)(x-q)=x^2-px-qx+pq
thus 0=4-2(p+q)-8
0=-4-2(p+q)
4=-2(p+q)
thus p+q=-2 & p= -2-q

subsituting
-8=pq=q(-2-q)=-2q-q^2
thus q^2+2q-8=0
s0(q+4)(q-2)=0
so if q=-4,p=2 and if q=2,p=-4.
either case, it satisfies the intercept at(2,0) and is sufficient.

similarly, you can solve for -2-p=q