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Math Expert V
Joined: 02 Sep 2009
Posts: 57279
Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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1
10 00:00

Difficulty:   55% (hard)

Question Stats: 66% (02:52) correct 34% (02:12) wrong based on 50 sessions

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Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

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Manager  P
Joined: 01 Feb 2017
Posts: 232
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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Equating points into given equation, we get
|a-π|=1
so a can be approx 4.14 or 2.14

|b-π|=1
so b can be approx 4.14 or 2.14

As a and b are distinct, difference a-b= +-2

Hence, |a-b|= 2

Ans C

Posted from my mobile device
Manager  S
Joined: 29 Dec 2018
Posts: 80
Location: India
WE: Marketing (Real Estate)
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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Dear Moderators Gladiator59, VeritasKarishma generis
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7
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Intern  B
Joined: 04 Jul 2019
Posts: 1
Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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Hey,

Put the Value of x=RootPi in the Equation, You'll get the Values of "y" ( Pi+1 and Pi-1).

Since There are only two solutions to the equation, so above values of "y" will be actually the values of "a" and "b". The order does not matter because of the absolute value sign.

Hope it Helps!
Manager  G
Joined: 19 Nov 2017
Posts: 220
Location: India
Schools: ISB
GMAT 1: 670 Q49 V32 GPA: 4
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides
$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2
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Vaibhav

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Manager  S
Joined: 29 Dec 2018
Posts: 80
Location: India
WE: Marketing (Real Estate)
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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vaibhav1221 wrote:
DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides

$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2

Shouldn't this be
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y - (x^2))^2 = 1$$
_________________
Keep your eyes on the prize: 750
Manager  G
Joined: 19 Nov 2017
Posts: 220
Location: India
Schools: ISB
GMAT 1: 670 Q49 V32 GPA: 4
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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DarkHorse2019 wrote:
vaibhav1221 wrote:
DarkHorse2019 wrote:
Dear Moderators Gladiator59, VeritasKarishma generis
Please can you help with the solution here?

Did not understand with the one provided by Shobhit7

DarkHorse2019, consider the following solution. I hope this will make sense.

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 = 2x^2 y + 1$$
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y + (x^2))^2 = 1$$
square rooting both sides

$$y + (x^2) = ±1$$

$$x^2 = y + 1$$ ...(1)
$$x^2 = y - 1$$ ...(2)

in equation (1)
when $$x = \sqrt{\pi}, y = (15/7)$$ (value of a or b)

in equation (2)
when $$x = \sqrt{\pi}, y = (29/7)$$ (value of b or a)

|a-b| =

|15/7 - 29/7| or |29/7 - 15/7| = |-14/7| or |14/7| = 2

Shouldn't this be
$$y^2 + (x^2)^2 - 2x^2 y = 1$$
$$(y - (x^2))^2 = 1$$

Yes, my bad. The procedure is fine though.
_________________
Regards,

Vaibhav

Sky is the limit. 800 is the limit.

~GMAC
Veritas Prep GMAT Instructor D
Joined: 16 Oct 2010
Posts: 9554
Location: Pune, India
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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2
1
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

Note that for both points, value of x is given to be $$\sqrt{\pi}$$. Plugging this in, let's get the values of y which are the values of a and b.

Graph:
$$y^2 + \pi^2 = 2\pi*y + 1$$

$$y^2 - 2\pi*y + \pi^2 - 1 = 0$$

$$[y - (\pi + 1)][y - (\pi - 1)] = 0$$

Since |x - y| is not 0 (No option is 0),
$$a = (\pi + 1)$$
$$b = (\pi - 1)$$
(or the other way around)

|a - b| = 2
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VP  P
Joined: 03 Jun 2019
Posts: 1081
Location: India
Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C
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Intern  B
Joined: 08 Dec 2018
Posts: 11
Location: India
Schools: ISB '20, IIMA PGPX"20
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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Kinshook wrote:
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C

Hi Kinshook

For - $$|a-\pi|=1$$ and $$|b-\pi|=1$$

How did you assume that |a-\pi| to be 1? It can take any value, right ?
Please suggest and help me with this.
Intern  B
Joined: 08 Dec 2018
Posts: 11
Location: India
Schools: ISB '20, IIMA PGPX"20
Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap  [#permalink]

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NamGup wrote:
Kinshook wrote:
Bunuel wrote:
Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph of $$y^2 + x^4 = 2x^2 y + 1$$. What is $$|a - b|$$ ?

(A) 1

(B) $$\frac{\pi}{2}$$

(C) 2

(D) $$\sqrt{1 + \pi}$$

(E) $$1 + \sqrt{\pi}$$

$$y^2 + x^4 = 2x^2 y + 1$$
$$y^2+x^4-2x^2y = 1$$
$$(y-x^2)^2 =1$$
$$|y-x^2| = 1$$

Points $$(\sqrt{\pi}, a)$$ and $$(\sqrt{\pi}, b)$$ are distinct points on the graph
=>$$|a-\pi|=1$$
& $$|b-\pi|=1$$
Since they are distinct points
Say$$a = \pi+1$$ and $$b =\pi-1$$
|a-b| =2 since a and b are interchangeable

IMO C

Hi Kinshook

For - $$|a-\pi|=1$$ and $$|b-\pi|=1$$

How did you assume that |a-\pi| to be 1? It can take any value, right ?
Please suggest and help me with this.

Ok.. I understood this point. you substituted the values in the equation. Re: Points (pi^(1/2), a) and (pi^(1/2), b) are distinct points on the grap   [#permalink] 24 Jul 2019, 06:09
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