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If x is positive and y is 1 more than the square of x, then

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by BTGmoderatorDC » Thu Oct 11, 2018 7:11 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?

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

OA D

Source: Princeton Review
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answer

by Byjus » Fri Oct 12, 2018 1:41 am
Hi,

It’s a straight forward question.

Given,

x is positive and y = x^2 + 1

y = x^2 +1

x^2 = y-1

x = sqrt(y-1)

So the answer is D.

OR

We can try numbers too.

If x = 1 then y = 2.

When we substitute y = 2 in the answer choices we should get “x” value as 1.

A. y^2 – 1 = 4 – 1 = 3. Eliminate.

B. y^2 + 1 = 4 + 1 = 5. Eliminate.

C. √(y)− 1 = √2 – 1 = not an integer. Eliminate.

D. √(y – 1) = √(2 – 1) = 1. Hold on.

E. √(y + 1) = √(2 + 1) = 3. Eliminate.

So the answer is D.
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by Scott@TargetTestPrep » Thu Oct 18, 2018 5:17 pm
BTGmoderatorDC wrote: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?

$$A.\ y^2\ -\ 1$$
$$B.\ y^2\ +\ 1$$
$$C.\ \sqrt{y}-\ 1$$
$$D.\ \sqrt{y\ -\ 1}$$
$$E.\ \sqrt{y\ +\ 1}$$
We can create the equation:

y = 1 + x^2

y - 1 = x^2

√(y - 1) = x

Answer: D

Scott Woodbury-Stewart
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