Q.Find the pairs of equal sets, if any, give reasons: A = {0}, B = {x : x > 15 and x < 5}, C = {x : x – 5 = 0 }, D = {x: x2 = 25}, E = {x : x is an integral positive root of the equation x2 – 2x –15 = 0}

Concept: Set Membership — two sets are equal if and only if they contain exactly the same elements.

Step 1: List each set explicitly.

$A = \{0\}$

$B = \{x : x > 15 \text{ and } x < 5\}$ — no number satisfies both, so $B = \varnothing$ (empty set).

$C = \{x : x - 5 = 0\} = \{5\}$

$D = \{x : x^2 = 25\} = \{-5, 5\}$

$E = \{x : x \text{ is an integral positive root of } x^2 - 2x - 15 = 0\}$ — solving: $x^2 - 2x - 15 = (x-5)(x+3)=0$, roots are $5$ and $-3$; only positive integral root is $5$, so $E = \{5\}$.

Step 2: Compare sets.

$C = \{5\}$ and $E = \{5\}$ — identical elements.

No other pair matches: $A$ has $0$, $B$ is empty, $D$ has two elements.

The question asks us to compare five sets and find which ones are exactly equal. Two sets are equal if and only if they contain precisely the same elements — no more, no less. So we need to write each set in roster form (listing its elements) and then compare.

Let’s go through each set one by one.

1. Set A = {0}

This is already in roster form. It contains exactly one element: the number 0.

2. Set B = {x : x > 15 and x < 5}

The condition says $x > 15$ and $x < 5$ simultaneously. No real number can be both greater than 15 and less than 5. So there is no element that satisfies this.

Hence, B is the empty set, written as $\emptyset$ or $\{\}$.

3. Set C = {x : x – 5 = 0}

Solve $x - 5 = 0 \implies x = 5$.

So C = $\{5\}$.

4. Set D = {x : $x^2 = 25$}

Solve $x^2 = 25$. This gives $x = 5$ or $x = -5$.

So D = $\{5, -5\}$.

5. Set E = {x : x is an integral positive root of the equation $x^2 – 2x – 15 = 0$}

First, solve the quadratic: $x^2 - 2x - 15 = 0$.

Factor: $(x - 5)(x + 3) = 0$.

So the roots are $x = 5$ and $x = -3$.

Now, the set only includes integral positive roots. Among the two roots, only $x = 5$ is positive (and it is an integer). So E = $\{5\}$.

Now compare:

• A = $\{0\}$ — not equal to any other set.
• B = $\emptyset$ — not equal to any other set (none of the others are empty).
• C = $\{5\}$ and E = $\{5\}$ — these are equal.
• D = $\{5, -5\}$ — has two elements, so not equal to any singleton set.

Method: Set-Builder to Roster Form Conversion

This method involves converting each set from set-builder notation into its explicit roster (list) form, then comparing the elements directly.

Steps

Step 1: Convert each set to roster form

• A = $\{0\}$

  Already in roster form: A = {0}

• B = $\{x : x > 15 \text{ and } x < 5\}$

  No number can be simultaneously greater than 15 and less than 5.

  So, B = ∅ (empty set)

• C = $\{x : x - 5 = 0\}$

  Solving: $x - 5 = 0 \implies x = 5$

  So, C = {5}

• D = $\{x : x^2 = 25\}$

  Solving: $x^2 = 25 \implies x = 5$ or $x = -5$

  So, D = {5, -5}

• E = $\{x : x \text{ is an integral positive root of } x^2 - 2x - 15 = 0\}$

  Solving: $x^2 - 2x - 15 = 0$

  Factor: $(x - 5)(x + 3) = 0$

  Roots: $x = 5$ or $x = -3$

  Only positive integral root: $x = 5$

  So, E = {5}

Step 2: Compare the sets

Step 3: Identify equal pairs

• C = E because both contain exactly the element 5

  Reason: $C = \{5\}$ and $E = \{5\}$ — same single element.

• No other pairs are equal

  • A has 0, no other set has 0.
  • B is empty, no other set is empty.
  • D has two elements {5, -5}, while C and E have only {5}.

Final Answer

The only equal pair is C and E.

Reason: Both sets contain exactly the element 5 (the solution of $x-5=0$ and the positive root of $x^2 - 2x - 15 = 0$).

Here are the common mistakes students make with this question, along with the reasoning to avoid them.

Mistake 1: Thinking B is an empty set because of a "contradiction"

• The error: Students see $x > 15$ and $x < 5$ and think "no number fits," so they write $B = \emptyset$. That part is correct. But then they mistakenly compare B to other sets as if $\emptyset$ is the same as $\{0\}$ or $\{5\}$.
• Why it’s wrong: The empty set has no elements. A set like $\{0\}$ has one element (the number 0). They are not equal.
• How to avoid: Always check the number of elements first. If one set has 0 elements and another has 1, they cannot be equal.

Mistake 2: Solving $x^2 = 25$ and forgetting the negative root

• The error: Students write $D = \{5\}$ because they only take $x = 5$ and ignore $x = -5$.
• Why it’s wrong: The equation $x^2 = 25$ has two solutions: $x = 5$ and $x = -5$. So $D = \{5, -5\}$.
• How to avoid: For any equation of the form $x^2 = a$ (where $a > 0$), always write $x = \pm \sqrt{a}$. List both values unless the problem restricts the domain (e.g., "positive integers").

Mistake 3: Solving $x^2 - 2x - 15 = 0$ and missing the "positive root" condition

• The error: Students solve the quadratic and get $x = 5$ and $x = -3$, then write $E = \{5, -3\}$.
• Why it’s wrong: The problem says "integral positive root" — so only $x = 5$ qualifies. $x = -3$ is an integer but not positive.
• How to avoid: Read the condition after solving. Circle or underline words like "positive," "negative," "natural," "integral." Then filter your solution set.

Mistake 4: Confusing $C = \{x : x - 5 = 0\}$ with $C = \{5\}$

• The error: This one is usually correct, but students sometimes write $C = \{0\}$ because they think "$x - 5 = 0$ means $x = 0$".
• Why it’s wrong: Solving $x - 5 = 0$ gives $x = 5$, not $x = 0$.
• How to avoid: Always solve the equation step-by-step: $x - 5 = 0 \implies x = 5$. Don't guess.

Mistake 5: Saying two sets are equal just because they share one element

• The error: For example, seeing that $A = \{0\}$ and $C = \{5\}$ both have one element, and thinking they are equal.
• Why it’s wrong: Two sets are equal only if they have exactly the same elements. $\{0\} \neq \{5\}$.
• How to avoid: Compare elements one by one. If even one element differs, the sets are not equal.

✓ Correct Answer for Reference

Equal pairs: $C = E$ (both are $\{5\}$).

No other pairs are equal.