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# If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin

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Math Expert
Joined: 02 Sep 2009
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If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

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06 Sep 2019, 04:36
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Difficulty:

15% (low)

Question Stats:

83% (01:25) correct 17% (01:59) wrong based on 58 sessions

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If (R, R^2 + 1) is the (x, y) coordinate of a point located on the line: y = 2x + 4, what can be the value of R?

A. –3.
B. 2.
C. 4.
D. 3.
E. 1.

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Joined: 18 Sep 2018
Posts: 64
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

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06 Sep 2019, 05:09
IMO D

A point on the line satisfies the equation. So substitute the point (R, R^2+1) into the equation.

We get R^2-2R-3 = 0.

Solving the above equation, we get R = 3 or -1.

From the options we have only 3.

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Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

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06 Sep 2019, 11:14
Bunuel wrote:
If (R, R^2 + 1) is the (x, y) coordinate of a point located on the line: y = 2x + 4, what can be the value of R?

A. –3.
B. 2.
C. 4.
D. 3.
E. 1.

equate for given x,y values
we get
r=3 & -1
IMO D; 3
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Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

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08 Sep 2019, 22:43
I'm not clear on the above solutions as the value i derived for R are -3 and 1 as opposed to the 3 and -1 derived by other users. here are my workings;
[quote][/quote]

y = 2x+4
R^2+1 = 2R+4
R^2 - 2R +1-4 = 0
R^2 - 2R -3 = 0

Thus R is -3 and 1 as only this provides the current sum of -2 and product of -3.

the other answers of 3 and - 1 result in a sum of 2 which does not agree with the equation.

Please clarify Bunuel ... kindly point out my error if any
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Joined: 18 Jul 2019
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Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin  [#permalink]

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24 Nov 2019, 01:09
OmotayoH wrote:
I'm not clear on the above solutions as the value i derived for R are -3 and 1 as opposed to the 3 and -1 derived by other users. here are my workings;
Quote:

y = 2x+4
R^2+1 = 2R+4
R^2 - 2R +1-4 = 0
R^2 - 2R -3 = 0

Thus R is -3 and 1 as only this provides the current sum of -2 and product of -3.

the other answers of 3 and - 1 result in a sum of 2 which does not agree with the equation.

Please clarify Bunuel ... kindly point out my error if any

R^2 - 2R -3 = 0
R^2-3R+R-3=0
R(R-3) + 1( R-3) = 0
(R-3)(R+1)=0
R=-3 /-1

Hope that helps
Re: If (R, R^2 + 1) is the (x, y) coordinate of a point located on the lin   [#permalink] 24 Nov 2019, 01:09
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