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

Question

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

(1)  q^2 > r

(2)  r^2 > q

Kurtosis · Answer

We have to check for the value of the discriminant, b^2 - 4ac

a = 1; b = 2q; c = r
b^2 - 4ac = 4q^2 - 4r

If 4q^2 - 4r > 0 --> There are 2 intercepts for x.
4q^2 - 4r = 0 --> One x intercept
4q^2 - 4r < 0 --> No real roots

St1: q^2 > r --> 4q^2 - 4r > 0 
Sufficient

St2: r^2 > q --> r can be positive or negative and we do not know whether q^2 is going to be greater or lesser than r.
Not Sufficient

Answer: A

matthewsmith_89 · Answer

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

profileusername · Answer

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

Bunuel 
 abhimahna

abhimahna · Answer

Hi  TheMastermind  ,

Here I go.

For this question, you need to understand two things:

1. If the equation is intersecting x axis, then y must be zero. Or I can say  x^2  + 2qx + r = 0
2. Roots of a quadratic equation( ax^2 + bx + c = 0 ) are given by the formula:

x =   /2a

and

x =   /2a

Now, in order to have real roots, the values inside the square root MUST be positive.

or I can say  b^2 - 4ac > = 0

Thus, when you make the similar equation with the question in hand,you will say you need  q^2 - r .

To get the real roots you will say  q^2 >= r .

This is what option A is doing.

if we have  q^2 > r , we will have 2 roots.

if we have  q^2 = r , we will have 1 root.

Hence, A is sufficient. It tells us that we have two roots or two intersecting points.

Does that make sense?

SaGa · Answer

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

hellosanthosh2k2 · Answer

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)

zanaik89 · Answer

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

and q^2=r will give one root?

Pls help thank u

abhimahna · Answer

Hi zanaik89 , Look at the general equations I mentioned: x = /2a and x = /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.

bumpbot · Answer

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

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