Problem 95 - Algebra for College Students

Question

Some of the product patterns can be used to do arithmetic computations mentally. For example, let's use the pattern \((a+b)^{2}=a^{2}+2 a b+b^{2}\) to compute \(31^{2}\) mentally. Your thought process should be " \(31^{2}=\) \((30+1)^{2}=30^{2}+2(30)(1)+1^{2}=961 . "\) Compute each of the following numbers mentally, and then check your answers. (a) \(21^{2}\) (b) \(41^{2}\) (c) \(71^{2}\) (d) \(32^{2}\) (e) \(52^{2}\) (f) \(82^{2}\)

Step-by-Step Solution

Answer

(a) 441, (b) 1681, (c) 5041, (d) 1024, (e) 2704, (f) 6724.

Step 1: Understanding the Pattern

The expansion pattern given is \((a+b)^2 = a^2 + 2ab + b^2\). This pattern allows us to decompose a number close to a multiple of 10 into two parts: \(a\) as the base multiple of 10 and \(b\) as the remainder. We will apply this pattern to square each provided number mentally.

Step 2: Applying the Pattern to 21^2

For \(21^2\), consider \(a=20\) and \(b=1\). Apply the pattern: \[(20+1)^2 = 20^2 + 2(20)(1) + 1^2\]Calculate each term:- \(20^2 = 400\)- \(2 \times 20 \times 1 = 40\)- \(1^2 = 1\)Add them together: \(400 + 40 + 1 = 441\).

Step 3: Applying the Pattern to 41^2

For \(41^2\), consider \(a=40\) and \(b=1\). Apply the pattern:\[(40+1)^2 = 40^2 + 2(40)(1) + 1^2\]Calculate each term:- \(40^2 = 1600\)- \(2 \times 40 \times 1 = 80\)- \(1^2 = 1\)Add them together: \(1600 + 80 + 1 = 1681\).

Step 4: Applying the Pattern to 71^2

For \(71^2\), consider \(a=70\) and \(b=1\). Apply the pattern:\[(70+1)^2 = 70^2 + 2(70)(1) + 1^2\]Calculate each term:- \(70^2 = 4900\)- \(2 \times 70 \times 1 = 140\)- \(1^2 = 1\)Add them together: \(4900 + 140 + 1 = 5041\).

Step 5: Applying the Pattern to 32^2

For \(32^2\), consider \(a=30\) and \(b=2\). Apply the pattern:\[(30+2)^2 = 30^2 + 2(30)(2) + 2^2\]Calculate each term:- \(30^2 = 900\)- \(2 \times 30 \times 2 = 120\)- \(2^2 = 4\)Add them together: \(900 + 120 + 4 = 1024\).

Step 6: Applying the Pattern to 52^2

For \(52^2\), consider \(a=50\) and \(b=2\). Apply the pattern:\[(50+2)^2 = 50^2 + 2(50)(2) + 2^2\]Calculate each term:- \(50^2 = 2500\)- \(2 \times 50 \times 2 = 200\)- \(2^2 = 4\)Add them together: \(2500 + 200 + 4 = 2704\).

Step 7: Applying the Pattern to 82^2

For \(82^2\), consider \(a=80\) and \(b=2\). Apply the pattern:\[(80+2)^2 = 80^2 + 2(80)(2) + 2^2\]Calculate each term:- \(80^2 = 6400\)- \(2 \times 80 \times 2 = 320\)- \(2^2 = 4\)Add them together: \(6400 + 320 + 4 = 6724\).

Key Concepts

Square of a Number

Squaring a number is a straightforward arithmetic process where you multiply the number by itself. In mathematical terms, squaring a number is represented as: \( x^2 \), where \( x \) is the number you want to square. By using mental arithmetic techniques, such as the decomposing method we learned in the provided exercise, you can make this process much faster and clearer.

For example, when we calculate \( 21^2 \), which is \( 21 \times 21 \), we can use the formula \((a + b)^2 = a^2 + 2ab + b^2\). Here, \( a \) and \( b \) are chosen such that they are easier to compute mentally. You identify numbers close to tens (like 20 in our example) and compute it quickly.

This approach makes mental calculation easier and more intuitive, especially with larger numbers or those not easily multiplied within a short timeframe.

Arithmetic Patterns

Arithmetic patterns are consistent, predictable series of numbers that allow calculations to be executed rapidly. Recognizing these patterns can greatly simplify complex calculations into more manageable portions. For example, the arithmetic pattern used in the exercise \((a + b)^2 = a^2 + 2ab + b^2\) reveals how numbers can be split into simpler components.

This specific pattern is helpful because

• It breaks down squaring of two-digit numbers.
• Places less cognitive load on mental calculations.
• Incorporates base 10 calculations, which are often simpler to understand.

When you apply these patterns, calculating squares like \( 41^2 \) becomes much less about memorizing outcomes and more about understanding the structure behind the numbers.

Mathematical Decomposition

Mathematical decomposition involves breaking numbers into parts that are easier to work with to solve problems more efficiently. By decomposing larger numbers into the sum of its parts, calculations can often be simplified. The exercise utilizes this concept by splitting numbers such as \( 71 \) into \( 70 + 1 \) to apply the pattern \((a + b)^2\). This works due to the simplicity of smaller number calculations.

The decomposition process typically includes:

• Identifying a nearby base (like a multiple of 10) to form one component.
• Distinguishing the leftover balance as the second component.
• Applying straightforward calculations to these components separately.

The beauty of decomposition is that it leverages familiarity with easier numbers to derive answers for more complex questions, creating a path to understand and solve numerical problems efficiently.