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

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

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New post 29 Mar 2017, 05:31
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  65% (hard)

Question Stats:

59% (01:12) correct 41% (01:34) wrong based on 150 sessions

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Re: How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 29 Mar 2017, 20:51
2
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
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Re: How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 30 Mar 2017, 18:28
I do not even know what the question is asking. Can someone please explain this to me.
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Re: How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 08 Sep 2017, 15:03
This question deserves a better explanation. Anyone who could offer some help?

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

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New post 09 Sep 2017, 02:06
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TheMastermind wrote:
This question deserves a better explanation. Anyone who could offer some help?


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

and

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

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?
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Re: How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 14 Sep 2017, 10:28
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
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How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 18 Oct 2017, 13:17
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)
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Re: How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 21 Oct 2017, 02:46
abhimahna wrote:
TheMastermind wrote:
This question deserves a better explanation. Anyone who could offer some help?


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

and

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

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?




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

and q^2=r will give one root?

Pls help thank u
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How many intersects with x-axis does y = x^2 + 2qx + r have ? [#permalink]

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New post 21 Oct 2017, 05:04
zanaik89 wrote:
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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How many intersects with x-axis does y = x^2 + 2qx + r have ?   [#permalink] 21 Oct 2017, 05:04
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