How many intersects with x-axis does y = x^2 + 2qx + r have ?

I do not even know what the question is asking. Can someone please explain this to me.

This question deserves a better explanation. Anyone who could offer some help?

Bunuel
abhimahna

Here are my 2 cents:
The given equation is quadratic. There can be 3 cases:
2 distinct roots- graph will intersect x axis twice.
2 equal roots- graph will intersect x axis once.
2 imaginary roots - graph will not intersect x axis.


To find the nature of roots we need to look at the discriminant of the equation:

D = \(4q^2 - 4r\)

Now looking at the condition we see that \(q^2 > r\) Hence \(q^2 - r>0\) or \(4q^2 - 4r > 0\)
Hence D>0 and graph will intersect X axis twice.

Condition 2:
\(r^2>q\). With this we cannot guess if \(q^2 - r>0 or <0\)
Not sufficient

The given equation points of intersection with x-axi => Basically the question is asking the number of roots of the given quadratic equation.

Case 1: Usually a quadratic equation has two roots, when it is of the form ax^2 + bx + c =0.=> no of points of intersection with x-axis is 2
Case 2: But a quadratic eqn can have both the roots equal, when it is of the form (x + a)^2 = 0 => no of points of intersection with x-axis is 1

Now, our job is to find out if the given equation falls into case2, if so no of point of intersection will be 1 else 2.
in the given equation, x^2+2qx+r = 0, if r = q^2, then equation becomes => x ^ 2 + 2qx + q ^2 => (x + q)^2 = 0 => case 2

so the question is r = q^2?

Statement 1:
q^2 > r => crystal clear sufficient that r != q^2, so the number of points of intersection is 2.

Statement 2:
r^2 > q => Not sufficient,
if q = 2, r = 4 (r = q^2), statement 2 satisfied => case 2, only one point of intersection
if q = 2, r = 5 (r != q^2), statement 2 satisfied => case 1, two points of intersection

So Answer (A)

How did u conclude that q^2 > r will give two roots?

and q^2=r will give one root?

Pls help thank u

Hi zanaik89 ,

Look at the general equations I mentioned:

x = \([ -b + \sqrt{b^2 - 4ac}]/2a\)

and

x = \([ -b - \sqrt{b^2 - 4ac}]/2a\)

Now, if \(\sqrt{b^2 - 4ac}\) = 0, we will have the same value of x for both the equations.

For this to be zero, I can say \(b^2 - 4ac\) needs to be 0. This means \(b^2\) = 4ac

Similarly, if I get \(\sqrt{b^2 - 4ac}\) > 0, I will get two different values of x. This means \(b^2\) > 4ac

Use the question in hand in a similar way, you will understand entire logic.

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