Algebra

[swerve]

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

[psarma]

Given y= \(x^2\) + 1
So \(x^2\) = y-1
x = \(\sqrt{y-1}\)

[Scott@TargetTestPrep]

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