Hi All,

can somebody please help me understand this? Here, when we get (x-a)(x-b) = 0, then why aren't the roots x=a and x=b ?

Then statement 1 will be sufficient ?

Please help!

VeritasKarishma Bunuel egmat chetan2u

This is a value question, which means that for a statement(s) to be sufficient you need to get single numerical value of the intersect points.

Hi Bunuel

I agree. However once we get x=a,b
I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1
x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device

Hi Bunuel

I agree. However once we get x=a,b
I used this info in st 1

a+b = -1

Since a=x and b=x

X+x = -1
x= -1/2

I know what I’m doing is not right but theoretically I cannot seem to understand why this approach is incorrect

Thanks

Posted from my mobile device

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

Answer (C).

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)

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)

X-intercepts of the function \(f(x)\) or in our case the function (graph) \(y=(x+a)(x+b)\) is the value(s) of \(x\) for \(y=0\). So basically the question asks to find the roots of quadratic equation \((x+a)(x+b)=0\).

\((x+a)(x+b)=0\) --> \(x^2+bx+ax+ab=0\) --> \(x^2+(a+b)x+ab=0\).

Statement (1) gives the value of \(a+b\), but we don't know the value of \(ab\) to solve the equation.

Statement (2) tells us the point of y-intercept, or the value of \(y\) when \(x=0\) --> \(y=(x+a)(x+b)=(0+a)(0+b)=ab=-6\). We know the value of \(ab\) but we don't know the value of \(a+b\) to solve the equation.

Together we know the values of both \(a+b\) and \(ab\), hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.


Answer: C.


Hope it's clear.