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Today, we're going to learn about finding the square of larger numbers without direct multiplication. For example, how would we find the square of 23?
Wouldnβt we just do 23 times 23?
Yes, but thatβs tedious for larger numbers! Instead, we can express 23 as 20 plus 3, then use the identity (a + b)Β². Can anyone repeat the formula?
(a + b)Β² = aΒ² + 2ab + bΒ²!
Great job! Now applying this to 23: (20 + 3)Β² becomes 20Β² + 2 * 20 * 3 + 3Β². What do we get?
Thatβs 400 + 120 + 9, which equals 529!
Exactly! Remember the key: breaking the number into parts can simplify calculations.
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Now letβs talk about a special case. How do we find the square of numbers ending in 5, like 35?
Do we still use the formula?
Absolutely! But first, observe the pattern: (a5)Β² = [a(a + 1) * 100] + 25. Can anyone explain this?
It seems like we multiply the a part by its next integer, then use that in hundreds and add 25.
Exactly! So for 35, what do we have?
3 * 4 equals 12, so itβs 1225!
Perfect! Make sure to remember this shortcut for squaring numbers ending in 5.
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Next, we will explore Pythagorean triplets. What do you know about them?
Arenβt they sets of three numbers where the square of one equals the sum of the squares of the other two?
Exactly! For example, 3, 4, and 5 is a triplet because 3Β² + 4Β² = 5Β². Can someone provide another example?
I think 6, 8, 10 is also one!
Right again! Does anyone know how we can generate more Pythagorean triplets?
I recall something about using a formula involving natural numbers?
Yes! For any natural number m > 1, (2m)Β² + (mΒ² - 1)Β² = (mΒ² + 1)Β². Letβs try to find one with the smallest member being 6.
We would set 2m = 6, so m is 3, leading to 6, 8, and 10 then!
Fantastic! Keep practicing generating those!
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The section illustrates techniques to find the square of larger numbers by using the identity
(a + b)Β² = aΒ² + 2ab + bΒ². It also explores special cases of squaring numbers ending in 5 and introduces the concept of Pythagorean triplets.
In this section, we explore how to find the square of larger numbers without multiplying them directly. The fundamental identity used is:
(a + b)Β² = aΒ² + 2ab + bΒ²
For instance, for the number 23, we can break it down into (20 + 3). Hence, the square is calculated as:
(20 + 3)Β² = 20Β² + 2 * 20 * 3 + 3Β² = 400 + 120 + 9 = 529.
We also consider special patterns for numbers ending in 5, such as (a5)Β² = [a(a + 1) * 100] + 25, and illustrate this with examples.
Lastly, the section briefly discusses Pythagorean triplets and introduces a formula for generating them using natural numbers, showcasing the relationship of squares between certain integer sets.
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Squares of small numbers like 3, 4, 5, 6, 7, ... etc. are easy to find. But can we find the square of 23 so quickly? The answer is not so easy and we may need to multiply 23 by 23.
Calculating the square of smaller numbers, such as 3, 4, 5, or 6, can be done easily because we memorize their squares. However, for larger numbers like 23, it can be tedious to rely on simple multiplication. This shows that although squaring is easy for small numbers, it becomes cumbersome for larger numbers.
Imagine if you regularly bake cookies. You have memorized the recipe for a dozen cookies (small number), but if someone asks you to bake 23 dozen, you might need to consult your notes or a calculator for accurate measurements.
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There is a way to find this without having to multiply 23 Γ 23. We know 23 = 20 + 3.
Therefore 23Β² = (20 + 3)Β² = 20(20 + 3) + 3(20 + 3) = 20Β² + 20 Γ 3 + 3 Γ 20 + 3Β² = 400 + 60 + 60 + 9 = 529.
To calculate the square of 23, we break it down into a sum, (20 + 3). This utilizes the algebraic identity for squaring a number: (a + b)Β² = aΒ² + 2ab + bΒ². By simplifying this, we arrive at the answer without direct multiplication, showing a pattern and structure in numbers that helps facilitate calculations.
Think of a recipe where you have to double ingredients. Instead of measuring out each ingredient twice, you can simply calculate 2 times the amount of a single serving, making it both easy and more efficient.
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Example 1: Find the square of the following numbers without actual multiplication. (i) 39 (ii) 42.
Solution: (i) 39Β² = (30 + 9)Β² = 30(30 + 9) + 9(30 + 9) = 30Β² + 30 Γ 9 + 9 Γ 30 + 9Β² = 900 + 270 + 270 + 81 = 1521.
(ii) 42Β² = (40 + 2)Β² = 40(40 + 2) + 2(40 + 2) = 40Β² + 40 Γ 2 + 2 Γ 40 + 2Β² = 1600 + 80 + 80 + 4 = 1764.
In this section, we calculate the squares of two numbers, 39 and 42, using the addition method we learned earlier. For 39, we broke it down as (30 + 9) and for 42, we used (40 + 2). This method allows us to avoid direct multiplication by using previously solved squares and the distributive property of multiplication.
Imagine organizing your wardrobe. Instead of buying clothes in bulk and estimating sizes, you can take careful measurements and find that a specific combination gives you the fit you need, just like combining smaller number segments makes squaring easier and more accurate.
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5.4.1 Other patterns in squares. Consider the following pattern: 25Β² = 625 = (2 Γ 3) hundreds + 25.
This chunk discusses recognizing patterns in numbers, particularly when the unit digit is 5. It shows that the square of any number ending in 5 can be found using a specific pattern: (a Γ (a + 1)) hundreds + 25, which greatly simplifies calculations.
Think of how certain flower arrangements can follow specific patterns based on the number of stems. Recognizing this pattern helps you create beautiful bouquets efficiently without having to calculate each combination from scratch.
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5.4.2 Pythagorean triplets. Consider the following (3Β² + 4Β² = 5Β²). The collection of numbers 3, 4, and 5 is known as a Pythagorean triplet.
A Pythagorean triplet consists of three positive integers a, b, and c, such that aΒ² + bΒ² = cΒ². This chunk presents basic examples like (3, 4, 5) and illustrates their relevance in finding relationships between squares of numbers. This concept is especially significant in geometry, as it helps in calculations regarding right triangles.
Consider a ladder leaning against a wall. The length of the ladder, the distance from the wall, and the height up the wall forms a right triangle, similar to how the Pythagorean triplet provides a method to determine these distances.
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Key Concepts
Algebraic Identity: Understanding the identity used for finding squares.
Squares of Numbers Ending in 5: A specific case that simplifies calculation.
Pythagorean Triplets: Sets of three numbers that satisfy the Pythagorean theorem.
See how the concepts apply in real-world scenarios to understand their practical implications.
To find the square of 39 without actual multiplication: (30 + 9)Β² = 30Β² + 2(30)(9) + 9Β² = 900 + 540 + 81 = 1521.
Finding the square of 95: (90 + 5)Β² = 90Β² + 2(90)(5) + 5Β² = 8100 + 900 + 25 = 9025.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
To find the square choose a and b; add their square for what you see!
A student wanted to simplify, found the square of 23 with a clever trick, turning it into a party of tens and ones that made it lick!
Pythagorean triplet: '2M' for 2m, then mΒ²-1, mΒ²+1 for fun!
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Square of a Number
Definition:
The result of multiplying a number by itself.
Term: Pythagorean Triplet
Definition:
A set of three positive integers a, b, and c, such that aΒ² + bΒ² = cΒ².
Term: Algebraic Identity
Definition:
An equality that holds for any values of its variables.