Algebra

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Algebra

by swerve » Thu Sep 24, 2020 1:35 pm

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If \(x\) is positive and \(y\) is \(1\) more than the square of \(x\), then what is the value of \(x\) in terms of \(y\)?

A. \(y^2-1\)
B. \(y^2+1\)
C. \(\sqrt{y}-1\)
D. \(\sqrt{y-1}\)
E. \(\sqrt{y+1}\)

The OA is D

Source: Princeton Review

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Re: Algebra

by psarma » Thu Sep 24, 2020 2:59 pm
Given y= \(x^2\) + 1
So \(x^2\) = y-1
x = \(\sqrt{y-1}\)

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Re: Algebra

by [email protected] » Thu Oct 01, 2020 9:56 am
swerve wrote:
Thu Sep 24, 2020 1:35 pm
If \(x\) is positive and \(y\) is \(1\) more than the square of \(x\), then what is the value of \(x\) in terms of \(y\)?

A. \(y^2-1\)
B. \(y^2+1\)
C. \(\sqrt{y}-1\)
D. \(\sqrt{y-1}\)
E. \(\sqrt{y+1}\)

The OA is D

Solution:

We are given that y = x^2 + 1. Thus:

x^2 = y - 1

x = ±√(y - 1)

However, since x is positive, then x = √(y - 1) only.

Answer: D

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