Math Puzzle: Solve For X In P = √((x−2)/(x+1))

Jan 19, 2026 by Andrew McMorgan 47 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of mathematics with a super engaging problem that'll get your brains buzzing. We've got this expression, P = √((x−2)/(x+1)), and a specific set of values for x: x ∈ {−3, −2, −1, 0, 1, 2, 3}. Your mission, should you choose to accept it, is to figure out for which of these x values our expression P will be Real, Non-Real, Rational, or Undefined. Ready to flex those mathematical muscles?

Understanding the Expression: What Makes P Tick?

Alright, let's break down the beast: P = √((x−2)/(x+1)). The core of this expression is the square root function. We all know that square roots can be a bit picky about what goes inside them. To get a real number out of a square root, the value inside (the radicand) must be zero or positive. If the radicand is negative, we step into the realm of non-real or imaginary numbers. This is our first major hurdle, guys. We need to ensure that the fraction (x−2)/(x+1) is greater than or equal to zero for P to be real. Conversely, if (x−2)/(x+1) is negative, then P will be non-real.

Now, let's talk about the fraction itself: (x−2)/(x+1). This fraction has its own set of rules. The biggest no-no in any fraction is dividing by zero. So, we absolutely must ensure that the denominator, (x+1), is not equal to zero. If (x+1) = 0, then the entire fraction becomes undefined, and consequently, our expression P is also undefined. This condition (x+1 ≠ 0, meaning x ≠ -1) is critical. For the values of x given, we need to watch out for x = -1 specifically.

Beyond being real, we also have the condition of P being rational. A rational number is essentially any number that can be expressed as a fraction p/q, where p and q are integers and q is not zero. For P = √((x−2)/(x+1)) to be rational, two things need to happen: first, P must be real (so (x−2)/(x+1) ≥ 0), and second, the value inside the square root, (x−2)/(x+1), must be a perfect square of a rational number. This means that (x−2)/(x+1) itself must be the square of a fraction like (a/b)², where a and b are integers. In simpler terms, when you calculate the value of (x−2)/(x+1) for a given x, the result must be a perfect square (like 0, 1, 4, 9, 1/4, 9/16, etc.). If (x−2)/(x+1) is a positive non-perfect square (like 2 or 3), then P will be an irrational real number. So, to summarize, for P to be rational, we need (x−2)/(x+1) to be a non-negative perfect square.

Let's get into the specific scenarios now. We have our set of x values: {−3, −2, −1, 0, 1, 2, 3}. We'll systematically plug each one in and analyze the outcome based on these rules. This is where the real fun begins, guys! We're going to build a table or a step-by-step analysis for each condition (Real, Non-Real, Rational, Undefined) to make sure we don't miss anything. Get your calculators ready, or better yet, let's do some mental math! This kind of problem-solving is exactly what makes mathematics so captivating – it’s like solving a puzzle where every piece has to fit perfectly.

a) Determining When P is Real

Alright, let's get down to business, guys! For P = √((x−2)/(x+1)) to be real, the expression inside the square root, which is our fraction (x−2)/(x+1), must be greater than or equal to zero. Mathematically, this means we need (x−2)/(x+1) ≥ 0. This inequality tells us that either both the numerator (x−2) and the denominator (x+1) are positive, or both are negative. We also have to remember our golden rule: the denominator (x+1) cannot be zero, so x ≠ -1. Let's examine our given set of x values: {−3, −2, −1, 0, 1, 2, 3} and see where this condition holds true.

First, let's consider the case where both (x−2) and (x+1) are positive. For (x+1) > 0, we need x > -1. For (x−2) > 0, we need x > 2. To satisfy both conditions simultaneously (x > -1 and x > 2), we must have x > 2. Looking at our set, the values that satisfy x > 2 are x = 3.

Now, let's consider the case where both (x−2) and (x+1) are negative. For (x+1) < 0, we need x < -1. For (x−2) < 0, we need x < 2. To satisfy both conditions simultaneously (x < -1 and x < 2), we must have x < -1. Looking at our set, the only value that satisfies x < -1 is x = -3 and x = -2.

What about when the numerator is zero? If (x−2) = 0, then x = 2. In this case, the fraction becomes 0/(2+1) = 0/3 = 0. Since 0 ≥ 0, this value of x makes P real. So, x = 2 is also a solution.

Let's systematically test each value from the set {−3, −2, −1, 0, 1, 2, 3}:

For x = −3: (x−2) = (−3−2) = −5. (x+1) = (−3+1) = −2. The fraction is (−5)/(−2) = 5/2. Since 5/2 ≥ 0, P is real.
For x = −2: (x−2) = (−2−2) = −4. (x+1) = (−2+1) = −1. The fraction is (−4)/(−1) = 4. Since 4 ≥ 0, P is real.
For x = −1: (x+1) = (−1+1) = 0. The denominator is zero, so the fraction is undefined. Therefore, P is undefined, not real.
For x = 0: (x−2) = (0−2) = −2. (x+1) = (0+1) = 1. The fraction is (−2)/1 = −2. Since −2 < 0, P is non-real.
For x = 1: (x−2) = (1−2) = −1. (x+1) = (1+1) = 2. The fraction is (−1)/2 = −1/2. Since −1/2 < 0, P is non-real.
For x = 2: (x−2) = (2−2) = 0. (x+1) = (2+1) = 3. The fraction is 0/3 = 0. Since 0 ≥ 0, P is real.
For x = 3: (x−2) = (3−2) = 1. (x+1) = (3+1) = 4. The fraction is 1/4. Since 1/4 ≥ 0, P is real.

So, the values of x from the set {−3, −2, −1, 0, 1, 2, 3} for which P is real are x = −3, x = −2, x = 2, and x = 3. These are the values where the radicand (x−2)/(x+1) is non-negative.

b) Identifying When P is Non-Real

Now let's tackle the non-real cases, guys! For P = √((x−2)/(x+1)) to be non-real, the expression inside the square root – that's our fraction (x−2)/(x+1) – must be negative. That is, we need (x−2)/(x+1) < 0. This inequality holds true when the numerator (x−2) and the denominator (x+1) have opposite signs. Remember, we're working with the set x ∈ {−3, −2, −1, 0, 1, 2, 3}. We've already done most of the heavy lifting in the previous section, so this should be a breeze!

Let's re-examine our calculations for each x value:

x = −3: Fraction is 5/2 (positive). P is real.
x = −2: Fraction is 4 (positive). P is real.
x = −1: Denominator is zero. P is undefined.
x = 0: (x−2) = −2, (x+1) = 1. Fraction is (−2)/1 = −2. Since −2 < 0, P is non-real.
x = 1: (x−2) = −1, (x+1) = 2. Fraction is (−1)/2 = −1/2. Since −1/2 < 0, P is non-real.
x = 2: Fraction is 0 (non-negative). P is real.
x = 3: Fraction is 1/4 (positive). P is real.

So, the values of x from the set {−3, −2, −1, 0, 1, 2, 3} for which P is non-real are x = 0 and x = 1. These are the specific points where the expression inside the square root dips into the negative zone.

c) Finding When P is Rational

Alright, let's level up and figure out when P = √((x−2)/(x+1)) is rational, you math whizzes! For P to be rational, two conditions must be met: first, P must be real (meaning (x−2)/(x+1) ≥ 0), and second, the value inside the square root, (x−2)/(x+1), must be a perfect square of a rational number. In simpler terms, the fraction (x−2)/(x+1) needs to evaluate to a number like 0, 1, 4, 9, 1/4, 9/16, etc. Let's go back to our trusty set {−3, −2, −1, 0, 1, 2, 3} and check our previous results.

We already know which x values make P real: x = −3, x = −2, x = 2, and x = 3. Now we just need to check if the fraction (x−2)/(x+1) for these x values results in a perfect square.

For x = −3: The fraction (x−2)/(x+1) evaluates to 5/2. Is 5/2 a perfect square of a rational number? No. So, P is real but irrational for x = -3.
For x = −2: The fraction (x−2)/(x+1) evaluates to 4. Is 4 a perfect square of a rational number? Yes! 4 = 2². Since P is real and the radicand is a perfect square, P = √4 = 2, which is rational. So, x = -2 is a solution!
For x = 2: The fraction (x−2)/(x+1) evaluates to 0. Is 0 a perfect square of a rational number? Yes! 0 = 0². Since P is real and the radicand is a perfect square, P = √0 = 0, which is rational. So, x = 2 is a solution!
For x = 3: The fraction (x−2)/(x+1) evaluates to 1/4. Is 1/4 a perfect square of a rational number? Yes! 1/4 = (1/2)². Since P is real and the radicand is a perfect square, P = √(1/4) = 1/2, which is rational. So, x = 3 is a solution!

Therefore, the values of x from the set {−3, −2, −1, 0, 1, 2, 3} for which P is rational are x = −2, x = 2, and x = 3. These are the sweet spots where P gives us a nice, clean rational number.

d) Pinpointing When P is Undefined

Finally, let's talk about when our expression P = √((x−2)/(x+1)) is undefined, guys. The primary way an expression can become undefined is through division by zero. In our case, this happens when the denominator of the fraction, (x+1), is equal to zero. If (x+1) = 0, then x = -1. When x = -1, the fraction (x−2)/(x+1) becomes (−1−2)/(-1+1) = −3/0, which is definitely undefined.

Let's check our given set of x values: {−3, −2, −1, 0, 1, 2, 3}. We need to find the value(s) of x that make the denominator zero.

For x = −3: Denominator is −2 (not zero).
For x = −2: Denominator is −1 (not zero).
For x = −1: Denominator is −1 + 1 = 0. Bingo! The denominator is zero here.
For x = 0: Denominator is 1 (not zero).
For x = 1: Denominator is 2 (not zero).
For x = 2: Denominator is 3 (not zero).
For x = 3: Denominator is 4 (not zero).

So, the only value of x from the set {−3, −2, −1, 0, 1, 2, 3} for which P is undefined is x = −1. This is the single point where our mathematical expression throws a fit and refuses to give us any value at all!

Summary of Results

Let's wrap this up with a clear summary for all you math enthusiasts:

P is Real for: x = −3, x = −2, x = 2, x = 3
P is Non-Real for: x = 0, x = 1
P is Rational for: x = −2, x = 2, x = 3
P is Undefined for: x = −1

There you have it, folks! We've successfully navigated the complexities of the expression P = √((x−2)/(x+1)) for the given set of x values. It's amazing how just a few constraints and definitions can lead to such distinct outcomes. Keep practicing these kinds of problems, and you'll become math wizards in no time! Don't forget to check back with Plastik Magazine for more brain-teasing challenges and cool math insights. Stay curious!