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If y is 1 more than the square of a positive number x, what is the val

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New post 25 Jul 2017, 01:26
1
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A
B
C
D
E

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If y is 1 more than the square of a positive number x, what is the value of x, in terms of y?

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

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New post 25 Jul 2017, 02:50
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From the question stem, \(y = x^2 + 1\)

Therefore, \(x^2 = y - 1\) or \(x=\sqrt{y-1}\)(Option A)

Method 2(Assuming numbers)

From the question stem, \(y = x^2 + 1\)
If x = 5, y = \(x^2 + 1 = 25+1 = 26\)

Only Option A(\(\sqrt{y-1} = \sqrt{26-1} = \sqrt{25} = 5\)) gives us the same value of x when we substitute the value of y.
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New post 26 Jul 2017, 16:31
MathRevolution wrote:
If y is 1 more than the square of a positive number x, what is the value of x, in terms of y?

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


We can create the following equation:

y = 1 + x^2

y - 1 = x^2

√(y - 1) = x

Answer: A
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New post 27 Jul 2017, 17:57
==> From \(y=1+x^2\) and \(x^2=y-1, x=\sqrt{y-1}\), the answer is A.

Answer: A
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Re: If y is 1 more than the square of a positive number x, what is the val   [#permalink] 27 Jul 2017, 17:57
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