Increments in Products
?In-text Question — Consider the multiplication of two numbers, say, 23 × 27.
1. By how much does the product increase if the first number (23) is increased by 1?
2. What if the second number (27) is increased by 1?
3. How about when both numbers are increased by 1?
Do you see a pattern that could help generalise our observations to the product of any two numbers?
The original product is \(23 \times 27 = 621\).
1. \(24 \times 27 = 648\).
The product increases by \(648 - 621 = 27\).
2. \(23 \times 28 = 644\).
The product increases by \(644 - 621 = 23\).
3. \(24 \times 28 = 672\).
The product increases by \(672 - 621 = 51\).
Pattern: the increases are 27, 23 and 51.
Note that \(51 = 27 + 23 + 1\).
In general, for the product \(ab\):
If \(a\) is increased by 1, the product increases by \(b\).
If \(b\) is increased by 1, the product increases by \(a\).
If both are increased by 1, the product increases by \(a + b + 1\).
?In-text Question — What would we get if we had expanded (a + 1) (b + 1) by first taking (b + 1) as a single term? Try it?
Take \((b + 1)\) as a single term.
\((a + 1)(b + 1) = a(b + 1) + 1(b + 1)\) (distributive property)
\(= ab + a + b + 1\)
This is the same as the result we got by taking \((a + 1)\) as a single term.
So the product increases by \(a + b + 1\).
?In-text Question — Will the product always increase? Find 3 examples where the product decreases.
This refers to \((a + 1)(b - 1)\), where one number is increased by 1 and the other is decreased by 1.
\((a + 1)(b - 1) = ab + b - a - 1\)
The change in the product is \(b - a - 1\).
The product decreases when \(b - a - 1 < 0\), that is, when \(b \le a\) (for whole numbers).
So the product does not always increase.
Examples where the product decreases:
\(a = 10, b = 3\): \(ab = 30\) and \((11)(2) = 22\).
It decreases by 8.
\(a = 5, b = 3\): \(ab = 15\) and \((6)(2) = 12\).
It decreases by 3.
\(a = 7, b = 7\): \(ab = 49\) and \((8)(6) = 48\).
It decreases by 1.
?In-text Question — What happens when a and b are negative integers? Check by substituting different values for a and b in each of the above cases. For example, a = –5, b = 8; a = –4, b = –5; etc.
We check the three results: increase \(a\) (when \(b\) goes up by 1), increase \(a + b + 1\) (both up by 1), and increase \(b - a - 1\) (\(a\) up by 1, \(b\) down by 1).
Case \(a = -5, b = 8\): \(ab = -40\).
\(b\) up by 1: \((-5)(9) = -45\).
Change \(= -45 - (-40) = -5 = a\).
Both up by 1: \((-4)(9) = -36\).
Change \(= -36 + 40 = 4\), and \(a + b + 1 = -5 + 8 + 1 = 4\).
\(a\) up, \(b\) down: \((-4)(7) = -28\).
Change \(= -28 + 40 = 12\), and \(b - a - 1 = 8 + 5 - 1 = 12\).
Case \(a = -4, b = -5\): \(ab = 20\).
\(b\) up by 1: \((-4)(-4) = 16\).
Change \(= 16 - 20 = -4 = a\).
Both up by 1: \((-3)(-4) = 12\).
Change \(= 12 - 20 = -8\), and \(a + b + 1 = -4 - 5 + 1 = -8\).
\(a\) up, \(b\) down: \((-3)(-6) = 18\).
Change \(= 18 - 20 = -2\), and \(b - a - 1 = -5 + 4 - 1 = -2\).
The distributive property holds for integers, so the same expressions work for negative values.
?In-text Question — a × a² = a³ (why?)
\(a^2 = a \times a\).
So \(a \times a^2 = a \times (a \times a)\).
\(= a \times a \times a\)
\(= a^3\) (three factors of \(a\))
?In-text Question — Use Identity 1 to find how the product changes when
(i) one number is decreased by 2 and the other increased by 3;
(ii) both numbers are decreased, one by 3 and the other by 4. Verify the answers by finding the products without converting the subtractions to additions.
Identity 1: \((a + m)(b + n) = ab + mb + an + mn\).
The change in the product is \(an + bm + mn\).
(i) \(m = -2\) and \(n = 3\).
Change \(= a(3) + b(-2) + (-2)(3)\)
\(= 3a - 2b - 6\)
Verify: \((a - 2)(b + 3) = a(b + 3) - 2(b + 3)\)
\(= ab + 3a - 2b - 6\)
So the change is \(3a - 2b - 6\).
(ii) \(m = -3\) and \(n = -4\).
Change \(= a(-4) + b(-3) + (-3)(-4)\)
\(= -4a - 3b + 12\)
Verify: \((a - 3)(b - 4) = a(b - 4) - 3(b - 4)\)
\(= ab - 4a - 3b + 12\)
So the change is \(-4a - 3b + 12\).
Numerical check with \(a = 5, b = 7\) (\(ab = 35\)): \((3)(10) = 30\), a change of \(-5\), and \(3(5) - 2(7) - 6 = -5\).
\((2)(3) = 6\), a change of \(-29\), and \(-4(5) - 3(7) + 12 = -29\).
(ii) The product changes by −4a − 3b + 12.
Figure it Out — Set 1 (Page 142)
Five questions from page 142 and page 143 of the textbook.
1Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.
In the example frame, the middle number is \(4 \times 6\).
Take \(p = 4\) and \(q = 6\).
The row above has the first factor \(p - 1 = 3\).
The row below has \(p + 1 = 5\).
The column to the left has the second factor \(q - 1 = 5\).
The column to the right has \(q + 1 = 7\).
So each number in the frame is a product of (a row factor) and (a column factor).
| q − 1 | q | q + 1 | |
|---|---|---|---|
| p − 1 | \((p-1)(q-1)\) | \((p-1)q\) | \((p-1)(q+1)\) |
| p | \(p(q-1)\) | \(pq\) | \(p(q+1)\) |
| p + 1 | \((p+1)(q-1)\) | \((p+1)q\) | \((p+1)(q+1)\) |
Check with \(p = 4, q = 6\): \((p-1)(q-1) = 3 \times 5\), \((p-1)q = 3 \times 6\), \((p-1)(q+1) = 3 \times 7\), and so on, matching the example frame.
2Expand the following products.
(i) (3 + u) (v – 3)
(ii) 2/3 (15 + 6a)
(iii) (10a + b) (10c + d)
(iv) (3 – x) (x – 6)
(v) (–5a + b) (c + d)
(vi) (5 + z) (y + 9)
(i) \((3 + u)(v - 3)\)
\(= 3(v - 3) + u(v - 3)\)
\(= 3v - 9 + uv - 3u\)
\(= uv - 3u + 3v - 9\)
(ii) \(\frac{2}{3}(15 + 6a)\)
\(= \frac{2}{3} \times 15 + \frac{2}{3} \times 6a\)
\(= 10 + 4a\)
(iii) \((10a + b)(10c + d)\)
\(= 10a(10c + d) + b(10c + d)\)
\(= 100ac + 10ad + 10bc + bd\)
(iv) \((3 - x)(x - 6)\)
\(= 3(x - 6) - x(x - 6)\)
\(= 3x - 18 - x^2 + 6x\)
\(= -x^2 + 9x - 18\)
(v) \((-5a + b)(c + d)\)
\(= -5a(c + d) + b(c + d)\)
\(= -5ac - 5ad + bc + bd\)
(vi) \((5 + z)(y + 9)\)
\(= 5(y + 9) + z(y + 9)\)
\(= 5y + 45 + yz + 9z\)
\(= yz + 5y + 9z + 45\)
(ii) 10 + 4a
(iii) 100ac + 10ad + 10bc + bd
(iv) −x² + 9x − 18
(v) −5ac − 5ad + bc + bd
(vi) yz + 5y + 9z + 45
3Math Talk — Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Let the numbers be \(a\) and \(b\).
We want \((a + 2)(b - 4) = ab\).
\((a + 2)(b - 4) = ab - 4a + 2b - 8\)
Setting this equal to \(ab\): \(-4a + 2b - 8 = 0\).
\(2b = 4a + 8\)
\(b = 2a + 4\)
So any pair with \(b = 2a + 4\) works.
Examples:
\(a = 1, b = 6\): \(1 \times 6 = 6\) and \(3 \times 2 = 6\).
\(a = 2, b = 8\): \(2 \times 8 = 16\) and \(4 \times 4 = 16\).
\(a = 3, b = 10\): \(3 \times 10 = 30\) and \(5 \times 6 = 30\).
4Expand
(i) (a + ab – 3b²) (4 + b), and
(ii) (4y + 7) (y + 11z – 3).
(i) \((a + ab - 3b^2)(4 + b)\)
\(= 4(a + ab - 3b^2) + b(a + ab - 3b^2)\)
\(= 4a + 4ab - 12b^2 + ab + ab^2 - 3b^3\)
\(= 4a + 5ab + ab^2 - 12b^2 - 3b^3\) (adding the like terms \(4ab\) and \(ab\))
(ii) \((4y + 7)(y + 11z - 3)\)
\(= 4y(y + 11z - 3) + 7(y + 11z - 3)\)
\(= 4y^2 + 44yz - 12y + 7y + 77z - 21\)
\(= 4y^2 + 44yz - 5y + 77z - 21\) (adding the like terms \(-12y\) and \(7y\))
(ii) 4y² + 44yz − 5y + 77z − 21
5Expand
(i) (a – b) (a + b),
(ii) (a – b) (a² + ab + b²) and
(iii) (a – b)(a³ + a²b + ab² + b³), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
(i) \((a - b)(a + b)\)
\(= a(a + b) - b(a + b)\)
\(= a^2 + ab - ab - b^2\)
\(= a^2 - b^2\)
(ii) \((a - b)(a^2 + ab + b^2)\)
\(= a(a^2 + ab + b^2) - b(a^2 + ab + b^2)\)
\(= a^3 + a^2b + ab^2 - a^2b - ab^2 - b^3\)
\(= a^3 - b^3\)
(iii) \((a - b)(a^3 + a^2b + ab^2 + b^3)\)
\(= a^4 + a^3b + a^2b^2 + ab^3 - a^3b - a^2b^2 - ab^3 - b^4\)
\(= a^4 - b^4\)
Pattern: \((a - b)\) times (the sum of terms from \(a^{n-1}\) down to \(b^{n-1}\), each power of \(a\) falling by 1 and each power of \(b\) rising by 1) gives \(a^n - b^n\).
Next identity: \((a - b)(a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 - b^5\).
Check by expanding:
\(a \times (a^4 + a^3b + a^2b^2 + ab^3 + b^4) = a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4\)
\(-b \times (a^4 + a^3b + a^2b^2 + ab^3 + b^4) = -a^4b - a^3b^2 - a^2b^3 - ab^4 - b^5\)
Adding, every middle term cancels in pairs, leaving \(a^5 - b^5\).
(ii) a³ − b³
(iii) a⁴ − b⁴
Next: (a − b)(a⁴ + a³b + a²b² + ab³ + b⁴) = a⁵ − b⁵
Fast Multiplication Using the Distributive Property
?In-text Question — Describe a general rule to multiply a number (of any number of digits) by 11 and write the product in one line.
Why it works: \(N \times 11 = N \times (10 + 1) = N \times 10 + N\).
For the 4-digit number \(dcba\):
\(dcba0 + dcba\), written one below the other with the digits lined up.
The result has the digits: \(d,\ (c + d),\ (b + c),\ (a + b),\ a\).
Rule:
Step 1: write the last digit of the number as the last digit of the product.
Step 2: each next digit (going left) is the sum of two neighbouring digits of the number.
The pair is the digit itself and the digit to its right.
Step 3: if a sum is 10 or more, write the units digit and carry 1 to the next sum on the left.
Step 4: the first digit of the number is written at the front (with any carry added to it).
Example: \(3874 \times 11\).
Last digit 4.
Then \(7 + 4 = 11\): write 1, carry 1.
Then \(8 + 7 + 1 = 16\): write 6, carry 1.
Then \(3 + 8 + 1 = 12\): write 2, carry 1.
Then \(3 + 1 = 4\).
The product is 42614.
MTMath Talk — Evaluate
(i) 94 × 11,
(ii) 495 × 11,
(iii) 3279 × 11,
(iv) 4791256 × 11.
(i) \(94 \times 11\): last digit 4.
Then \(9 + 4 = 13\): write 3, carry 1.
First digit \(9 + 1 = 10\).
The product is 1034.
(ii) \(495 \times 11\): last digit 5.
Then \(9 + 5 = 14\): write 4, carry 1.
Then \(4 + 9 + 1 = 14\): write 4, carry 1.
First digit \(4 + 1 = 5\).
The product is 5445.
(iii) \(3279 \times 11\): last digit 9.
Then \(7 + 9 = 16\): write 6, carry 1.
Then \(2 + 7 + 1 = 10\): write 0, carry 1.
Then \(3 + 2 + 1 = 6\).
First digit 3.
The product is 36069.
(iv) \(4791256 \times 11\): last digit 6.
Then \(5 + 6 = 11\): write 1, carry 1.
Then \(2 + 5 + 1 = 8\).
Then \(1 + 2 = 3\).
Then \(9 + 1 = 10\): write 0, carry 1.
Then \(7 + 9 + 1 = 17\): write 7, carry 1.
Then \(4 + 7 + 1 = 12\): write 2, carry 1.
First digit \(4 + 1 = 5\).
The product is 52703816.
(ii) 5445
(iii) 36069
(iv) 52703816
?In-text Question — Can we come up with a similar rule for multiplying a number by 101? Multiply 3874 by 101. Let us take a 4-digit number dcba. dcba × 101 = dcba × (100 + 1) = dcba × 100 + dcba. Use this to multiply 3874 × 101 in one line. What could be a general rule to multiply a number by 101 and write the product in one line? Extend this rule for multiplication by 1001, 10001, …
For 101: \(dcba \times 101 = dcba00 + dcba\).
Adding with the digits lined up: \(d,\ c,\ (b + d),\ (a + c),\ b,\ a\).
\(3874 \times 101\): \(d = 3, c = 8, b = 7, a = 4\).
\(3,\ 8,\ (7 + 3),\ (4 + 8),\ 7,\ 4\) gives \(3,\ 8,\ 10,\ 12,\ 7,\ 4\).
Carrying: \(12\) gives 2 and carry 1, so \(10 + 1 = 11\) gives 1 and carry 1, so \(8 + 1 = 9\).
The product is 391274.
Rule for 101: write the last two digits of the number.
Then each next digit is the sum of the digit and the digit two places to its right.
The first two digits are written as they are (plus carry).
For 1001: \(dcba \times 1001 = dcba000 + dcba\), which gives \(d,\ c,\ b,\ (a + d),\ c,\ b,\ a\).
For 10001: \(dcba \times 10001 = dcba0000 + dcba\), which gives \(d,\ c,\ b,\ a,\ d,\ c,\ b,\ a\).
The number is simply written twice.
General rule: for \(10^k + 1\), add the number to a copy of itself shifted \(k\) places to the left.
MTMath Talk — Use this to find
(i) 89 × 101,
(ii) 949 × 101,
(iii) 265831 × 1001,
(iv) 1111 × 1001,
(v) 9734 × 99 and
(vi) 23478 × 999.
(i) \(89 \times 101 = 8900 + 89 = 8989\).
(ii) \(949 \times 101 = 94900 + 949 = 95849\).
(iii) \(265831 \times 1001 = 265831000 + 265831\).
\(= 266096831\)
(iv) \(1111 \times 1001 = 1111000 + 1111\).
\(= 1112111\)
(v) \(9734 \times 99 = 9734(100 - 1)\).
\(= 973400 - 9734\)
\(= 963666\)
(vi) \(23478 \times 999 = 23478(1000 - 1)\).
\(= 23478000 - 23478\)
\(= 23454522\)
(ii) 95849
(iii) 266096831
(iv) 1112111
(v) 963666
(vi) 23454522
Square of the Sum or Difference of Two Numbers
?In-text Question — What if we write 65² as (30 + 35)² or (52 + 13)²? Draw the figures and check the area that you get.
Using \((30 + 35)^2\): a square of side 65 splits into a square of side 30, a square of side 35 and two \(30 \times 35\) rectangles.
\(30^2 = 900\)
\(35^2 = 1225\)
\(2 \times (30 \times 35) = 2 \times 1050 = 2100\)
Area \(= 900 + 1225 + 2100 = 4225\).
Using \((52 + 13)^2\): a square of side 65 splits into a square of side 52, a square of side 13 and two \(52 \times 13\) rectangles.
\(52^2 = 2704\)
\(13^2 = 169\)
\(2 \times (52 \times 13) = 2 \times 676 = 1352\)
Area \(= 2704 + 169 + 1352 = 4225\).
MTMath Talk — If a and b are any two integers, is (a + b)² always greater than a² + b²? If not, when is it greater?
\((a + b)^2 = a^2 + 2ab + b^2\).
So \((a + b)^2 - (a^2 + b^2) = 2ab\).
If \(ab > 0\) (both positive or both negative): \(2ab > 0\), so \((a + b)^2\) is greater.
Example: \(a = 3, b = 4\): \(49 > 25\).
And \(a = -3, b = -4\): \(49 > 25\).
If \(ab = 0\) (one of them is 0): \(2ab = 0\), so the two are equal.
Example: \(a = 0, b = 5\): \(25 = 25\).
If \(ab < 0\) (one positive, one negative): \(2ab < 0\), so \((a + b)^2\) is smaller.
Example: \(a = 3, b = -4\): \((-1)^2 = 1 < 25\).
?In-text Question — Use Identity 1A to find the values of 104², 37². (Hint: Decompose 104 and 37 into sums or differences of numbers whose squares are easy to compute.)
Identity 1A: \((a + b)^2 = a^2 + 2ab + b^2\).
\(104^2\): \(104 = 100 + 4\).
\(104^2 = 100^2 + 4^2 + 2 \times 100 \times 4\)
\(= 10000 + 16 + 800\)
\(= 10816\)
\(37^2\): \(37 = 30 + 7\).
\(37^2 = 30^2 + 7^2 + 2 \times 30 \times 7\)
\(= 900 + 49 + 420\)
\(= 1369\)
?In-text Question — Use Identity 1A to write the expressions for the following.
(i) (m + 3)²
(ii) (6 + p)²
(i) \((m + 3)^2 = m^2 + 3^2 + 2 \times m \times 3\)
\(= m^2 + 9 + 6m\)
\(= m^2 + 6m + 9\)
(ii) \((6 + p)^2 = 6^2 + p^2 + 2 \times 6 \times p\)
\(= 36 + p^2 + 12p\)
\(= p^2 + 12p + 36\)
(ii) p² + 12p + 36
?In-text Question — Expand (3j + 2k)² using both the identity and by applying the distributive property.
Using the identity \((a + b)^2 = a^2 + 2ab + b^2\) with \(a = 3j, b = 2k\):
\((3j + 2k)^2 = (3j)^2 + (2k)^2 + 2 \times 3j \times 2k\)
\(= 9j^2 + 4k^2 + 12jk\)
\(= 9j^2 + 12jk + 4k^2\)
Using the distributive property:
\((3j + 2k)(3j + 2k) = 3j(3j + 2k) + 2k(3j + 2k)\)
\(= 9j^2 + 6jk + 6jk + 4k^2\)
\(= 9j^2 + 12jk + 4k^2\)
?In-text Question — Find the general expansion of (a – b)² using geometry, as we did for 55².
Take a square of side \(a\).
Its area is \(a^2\).
Inside it, draw a square of side \(a - b\) in one corner.
Its area is \((a - b)^2\).
The square of side \(a - b\) is what remains after removing the two strips of width \(b\).
Each strip is a rectangle \(a \times b\) (area \(ab\)).
Removing both strips removes the small corner square \(b \times b\) twice.
So we add \(b^2\) back once.
\((a - b)^2 = a^2 - ab - ab + b^2\)
\(= a^2 - 2ab + b^2\)
?In-text Question — Use the identity (a – b)² to find the values of (a) 99² and (b) 58².
(a) \(99 = 100 - 1\)
\(99^2 = 100^2 - 2 \times 100 \times 1 + 1^2\)
\(= 10000 - 200 + 1\)
\(= 9801\)
(b) \(58 = 60 - 2\)
\(58^2 = 60^2 - 2 \times 60 \times 2 + 2^2\)
\(= 3600 - 240 + 4\)
\(= 3364\)
?In-text Question — Expand the following using both Identity 1B and by applying the distributive property.
(i) (b – 6)²
(ii) (–2a + 3)²
(iii) (7y – 3/(4z))²
(i) \((b - 6)^2\)
Identity 1B: \(b^2 + 6^2 - 2 \times b \times 6 = b^2 + 36 - 12b = b^2 - 12b + 36\)
Distributive property: \((b - 6)(b - 6) = b(b - 6) - 6(b - 6)\)
\(= b^2 - 6b - 6b + 36\)
\(= b^2 - 12b + 36\)
(ii) \((-2a + 3)^2\)
Write \(-2a + 3\) as \(3 - 2a\).
Identity 1B with first term 3 and second term \(2a\): \(3^2 + (2a)^2 - 2 \times 3 \times 2a\)
\(= 4a^2 + 9 - 12a\)
\(= 4a^2 - 12a + 9\)
Distributive property: \((-2a + 3)(-2a + 3) = -2a(-2a + 3) + 3(-2a + 3)\)
\(= 4a^2 - 6a - 6a + 9\)
\(= 4a^2 - 12a + 9\)
(iii) \(\left(7y - \frac{3}{4z}\right)^2\)
Identity 1B: \((7y)^2 + \left(\frac{3}{4z}\right)^2 - 2 \times 7y \times \frac{3}{4z}\)
\(= 49y^2 + \frac{9}{16z^2} - \frac{42y}{4z}\)
\(= 49y^2 - \frac{21y}{2z} + \frac{9}{16z^2}\)
Distributive property: \(7y\left(7y - \frac{3}{4z}\right) - \frac{3}{4z}\left(7y - \frac{3}{4z}\right)\)
\(= 49y^2 - \frac{21y}{4z} - \frac{21y}{4z} + \frac{9}{16z^2}\)
\(= 49y^2 - \frac{21y}{2z} + \frac{9}{16z^2}\)
(ii) 4a² − 12a + 9
(iii) 49y² − 21y/(2z) + 9/(16z²)
Investigating Patterns
?In-text Question — Use Identity 1C to calculate 98 × 102, and 45 × 55.
Identity 1C: \((a + b)(a - b) = a^2 - b^2\).
\(98 \times 102\): 98 and 102 are 2 below and 2 above 100.
\(98 \times 102 = (100 - 2)(100 + 2)\)
\(= 100^2 - 2^2\)
\(= 10000 - 4\)
\(= 9996\)
\(45 \times 55\): 45 and 55 are 5 below and 5 above 50.
\(45 \times 55 = (50 - 5)(50 + 5)\)
\(= 50^2 - 5^2\)
\(= 2500 - 25\)
\(= 2475\)
TTTry This — Show that (a + b) × (a – b) = a² – b² geometrically. Hint: a square of side a with a square of side b cut from a corner. What do we get when this piece is moved?
Take a square of side \(a\) and cut out a small square of side \(b\) from one corner.
The area left is \(a^2 - b^2\).
Cut the left part of the top strip (width \(a - b\), height \(b\)) and move it to the right side of the bottom block.
The bottom block is a rectangle of length \(a\) and height \(a - b\).
The moved piece is \(b\) wide (after turning it) and \(a - b\) high.
It fits next to the block on the right.
The new shape is a rectangle of length \(a + b\) and height \(a - b\).
Its area is \((a + b)(a - b)\).
The area did not change when the piece was moved.
?In-text Question — Sridharacharya (750 CE) gave an interesting method to quickly compute the squares of numbers using Identity 1C! Consider the following modified form of this identity — a² = (a + b) (a – b) + b². Why is this identity true?
By Identity 1C, \((a + b)(a - b) = a^2 - b^2\).
Add \(b^2\) to both sides.
\((a + b)(a - b) + b^2 = a^2 - b^2 + b^2\)
\(= a^2\)
Example: \(31^2\): take \(b = 1\), so \(a + b = 32\) and \(a - b = 30\).
\(31^2 = 32 \times 30 + 1^2 = 961\).
Figure it Out — Set 2 (Page 149)
Four questions from page 149 of the textbook.
1Which is greater: (a – b)² or (b – a)²? Justify your answer.
\((b - a) = -(a - b)\).
So \((b - a)^2 = [-(a - b)]^2 = (-1)^2(a - b)^2 = (a - b)^2\).
By expanding:
\((a - b)^2 = a^2 - 2ab + b^2\)
\((b - a)^2 = b^2 - 2ba + a^2 = a^2 - 2ab + b^2\)
Both expressions are the same.
Example: \(a = 7, b = 3\): \((4)^2 = 16\) and \((-4)^2 = 16\).
2Express 100 as the difference of two squares.
We want \(a^2 - b^2 = 100\).
By Identity 1C, \(a^2 - b^2 = (a + b)(a - b)\).
So we need \((a + b)(a - b) = 100\).
Both \((a + b)\) and \((a - b)\) must be even (they have the same parity, and 100 is even).
Take \(a + b = 50\) and \(a - b = 2\).
Adding: \(2a = 52\), so \(a = 26\).
Then \(b = 24\).
Check: \(26^2 - 24^2 = 676 - 576 = 100\).
(The pair \(a + b = 10\), \(a - b = 10\) gives \(a = 10, b = 0\), that is, \(100 = 10^2 - 0^2\).)
3Math Talk — Find 406², 72², 145², 1097², and 124² using the identities you have learnt so far.
\(406^2\): \(406 = 400 + 6\).
\(= 400^2 + 6^2 + 2 \times 400 \times 6 = 160000 + 36 + 4800\)
\(= 164836\)
\(72^2\): \(72 = 70 + 2\).
\(= 70^2 + 2^2 + 2 \times 70 \times 2 = 4900 + 4 + 280\)
\(= 5184\)
\(145^2\): \(145 = 150 - 5\).
\(= 150^2 + 5^2 - 2 \times 150 \times 5 = 22500 + 25 - 1500\)
\(= 21025\)
\(1097^2\): \(1097 = 1100 - 3\).
\(= 1100^2 + 3^2 - 2 \times 1100 \times 3 = 1210000 + 9 - 6600\)
\(= 1203409\)
\(124^2\): \(124 = 125 - 1\).
\(= 125^2 + 1^2 - 2 \times 125 \times 1 = 15625 + 1 - 250\)
\(= 15376\)
4Do Patterns 1 and 2 hold only for counting numbers? Do they hold for negative integers as well? What about fractions? Justify your answer.
Pattern 1: \(2(a^2 + b^2) = (a + b)^2 + (a - b)^2\).
Pattern 2: \(a^2 - b^2 = (a + b)(a - b)\).
Why they hold for all numbers: both were proved using only the distributive property, the commutative property and the rules for multiplying signs.
These hold for all integers and all fractions.
So both patterns hold for every kind of number.
Pattern 1, negative integers: \(a = -3, b = 2\).
LHS \(= 2(9 + 4) = 26\).
RHS \(= (-1)^2 + (-5)^2 = 1 + 25 = 26\).
Pattern 1, fractions: \(a = \frac{1}{2}, b = \frac{1}{3}\).
LHS \(= 2\left(\frac{1}{4} + \frac{1}{9}\right) = 2 \times \frac{13}{36} = \frac{13}{18}\).
RHS \(= \left(\frac{5}{6}\right)^2 + \left(\frac{1}{6}\right)^2 = \frac{25}{36} + \frac{1}{36} = \frac{13}{18}\).
Pattern 2, negative integers: \(a = -3, b = 2\).
LHS \(= 9 - 4 = 5\).
RHS \(= (-1)(-5) = 5\).
Pattern 2, fractions: \(a = \frac{3}{2}, b = \frac{1}{2}\).
LHS \(= \frac{9}{4} - \frac{1}{4} = 2\).
RHS \(= 2 \times 1 = 2\).
Mind the Mistake, Mend the Mistake
?In-text Question — We have expanded and simplified some algebraic expressions below to their simplest forms.
(i) Check each of the simplifications and see if there is a mistake.
(ii) If there is a mistake, try to explain what could have gone wrong.
(iii) Then write the correct expression.No. Simplification 1 \(-3p(-5p + 2q)\)
\(= -3p + 5p - 2q\)
\(= p - 2q\) 2 \(2(x - 1) + 3(x + 4)\)
\(= 2x - 1 + 3x + 4\)
\(= 5x + 3\) 3 \(y + 2(y + 2)\)
\(= (y + 2)^2\)
\(= y^2 + 4y + 4\) 4 \((5m + 6n)^2\)
\(= 25m^2 + 36n^2\) 5 \((-q + 2)^2\)
\(= q^2 - 4q + 4\) 6 \(3a(2b \times 3c)\)
\(= 6ab \times 9ac\)
\(= 54a^2bc\) 7 \(\frac{1}{2}(10s - 6) + 3\)
\(= 5s - 3 + 3\)
\(= 5s\) 8 \(5w^2 + 6w\)
\(= 11w^2\) 9 \(2a^3 + 3a^3 + 6a^2b + 6ab^2\)
\(= 5a^3 + 12a^2b^2\) 10 \((x + 2)(x + 5)\)
\(= (x + 2)x + (x + 2)5\)
\(= x^2 + 2x + 5x + 10\)
\(= x^2 + 7x + 10\) 11 \((a + 2)(b + 4)\)
\(= ab + 8\) 12 \(ab^2 + a^2b + a^2b^2\)
\(= ab(a + b + ab)\)
(i) Check each of the simplifications and see if there is a mistake.
(ii) If there is a mistake, try to explain what could have gone wrong.
(iii) Then write the correct expression.
| No. | Simplification |
|---|---|
| 1 | \(-3p(-5p + 2q)\) \(= -3p + 5p - 2q\) \(= p - 2q\) |
| 2 | \(2(x - 1) + 3(x + 4)\) \(= 2x - 1 + 3x + 4\) \(= 5x + 3\) |
| 3 | \(y + 2(y + 2)\) \(= (y + 2)^2\) \(= y^2 + 4y + 4\) |
| 4 | \((5m + 6n)^2\) \(= 25m^2 + 36n^2\) |
| 5 | \((-q + 2)^2\) \(= q^2 - 4q + 4\) |
| 6 | \(3a(2b \times 3c)\) \(= 6ab \times 9ac\) \(= 54a^2bc\) |
| 7 | \(\frac{1}{2}(10s - 6) + 3\) \(= 5s - 3 + 3\) \(= 5s\) |
| 8 | \(5w^2 + 6w\) \(= 11w^2\) |
| 9 | \(2a^3 + 3a^3 + 6a^2b + 6ab^2\) \(= 5a^3 + 12a^2b^2\) |
| 10 | \((x + 2)(x + 5)\) \(= (x + 2)x + (x + 2)5\) \(= x^2 + 2x + 5x + 10\) \(= x^2 + 7x + 10\) |
| 11 | \((a + 2)(b + 4)\) \(= ab + 8\) |
| 12 | \(ab^2 + a^2b + a^2b^2\) \(= ab(a + b + ab)\) |
| No. | Simplification | Mistake | Corrected statement |
|---|---|---|---|
| 1 | \(-3p(-5p + 2q)\) \(= -3p + 5p - 2q\) \(= p - 2q\) | \(-3p\) was not multiplied into each term. \(-3p \times (-5p)\) was written as \(-3p + 5p\) and \(-3p \times 2q\) as \(-2q\). Also \(-3p + 5p\) is \(2p\), not \(p\). | \(-3p \times (-5p) = 15p^2\) \(-3p \times 2q = -6pq\) \(-3p(-5p + 2q) = 15p^2 - 6pq\) |
| 2 | \(2(x - 1) + 3(x + 4)\) \(= 2x - 1 + 3x + 4\) \(= 5x + 3\) | The multipliers 2 and 3 were applied to the first term of each bracket only. | \(2(x - 1) = 2x - 2\) \(3(x + 4) = 3x + 12\) \(2x - 2 + 3x + 12 = 5x + 10\) |
| 3 | \(y + 2(y + 2)\) \(= (y + 2)^2\) \(= y^2 + 4y + 4\) | Operations were done in the wrong order. Multiplication comes first, so only \((y + 2)\) is multiplied by 2. | \(y + 2(y + 2)\) \(= y + 2y + 4\) \(= 3y + 4\) |
| 4 | \((5m + 6n)^2\) \(= 25m^2 + 36n^2\) | The middle term was left out. \((a + b)^2\) is not \(a^2 + b^2\). | \((5m + 6n)^2\) \(= (5m)^2 + 2(5m)(6n) + (6n)^2\) \(= 25m^2 + 60mn + 36n^2\) |
| 5 | \((-q + 2)^2\) \(= q^2 - 4q + 4\) | No mistake. | \((-q + 2)^2\) \(= q^2 - 4q + 4\) |
| 6 | \(3a(2b \times 3c)\) \(= 6ab \times 9ac\) \(= 54a^2bc\) | \((2b \times 3c)\) is just a product, \(6bc\). \(3a\) was wrongly multiplied into both factors. | \(3a(2b \times 3c) = 3a \times 6bc\) \(= 18abc\) |
| 7 | \(\frac{1}{2}(10s - 6) + 3\) \(= 5s - 3 + 3\) \(= 5s\) | No mistake. | \(\frac{1}{2}(10s - 6) + 3 = 5s\) |
| 8 | \(5w^2 + 6w\) \(= 11w^2\) | \(5w^2\) and \(6w\) are not like terms (\(w^2\) and \(w\) differ), so they cannot be added. | \(5w^2 + 6w\) cannot be simplified further. It can be written as \(w(5w + 6)\). |
| 9 | \(2a^3 + 3a^3 + 6a^2b + 6ab^2\) \(= 5a^3 + 12a^2b^2\) | \(6a^2b\) and \(6ab^2\) are not like terms, so they cannot become \(12a^2b^2\). Only \(2a^3 + 3a^3 = 5a^3\) can be added. | \(2a^3 + 3a^3 + 6a^2b + 6ab^2\) \(= 5a^3 + 6a^2b + 6ab^2\) |
| 10 | \((x + 2)(x + 5)\) \(= (x + 2)x + (x + 2)5\) \(= x^2 + 2x + 5x + 10\) \(= x^2 + 7x + 10\) | No mistake. | \((x + 2)(x + 5)\) \(= x^2 + 7x + 10\) |
| 11 | \((a + 2)(b + 4)\) \(= ab + 8\) | Only the first terms and the last terms were multiplied. Every term of one bracket must be multiplied by every term of the other. | \((a + 2)(b + 4)\) \(= ab + 4a + 2b + 8\) |
| 12 | \(ab^2 + a^2b + a^2b^2\) \(= ab(a + b + ab)\) | No mistake. | \(ab(a + b + ab)\) \(= a^2b + ab^2 + a^2b^2\) |
This Way or That Way, All Ways Lead to the Bay
MTMath Talk — Observe the pattern in the figure below. Draw the next figure in the sequence. How many circles does it have? How many total circles are there in Step 10? Write an expression for the number of circles in Step k.
Steps 1, 2, 3 have 3, 8 and 15 circles.
Each step is a square of circles with one corner circle missing.
Step 1 is \(2 \times 2\) minus 1.
Step 2 is \(3 \times 3\) minus 1.
Step 3 is \(4 \times 4\) minus 1.
So Step \(k\) is \((k + 1) \times (k + 1)\) minus 1.
Next figure (Step 4): a \(5 \times 5\) square with one circle missing.
Step 4 has \(5^2 - 1 = 24\) circles.
Step 10: \(11^2 - 1 = 121 - 1 = 120\) circles.
Step \(k\): \((k + 1)^2 - 1 = k^2 + 2k + 1 - 1 = k^2 + 2k\).
?In-text Question — Use this formula to find the number of circles in Step 15.
The number of circles in Step \(k\) is \(k^2 + 2k\).
For \(k = 15\): \(15^2 + 2 \times 15 = 225 + 30\).
\(= 255\)
Check: \((15 + 1)^2 - 1 = 256 - 1 = 255\).
MTMath Talk — Consider the pattern made of square tiles in the picture below. How many square tiles are there in each figure? How many are there in Step 4 of the sequence? What about Step 10? Write an algebraic expression for the number of tiles in Step n. Share your methods with the class. Can you find more than one method to arrive at the answer?
Tiles in each figure: Step 1 has 8 tiles, Step 2 has 12 tiles and Step 3 has 16 tiles.
Method 1 (big square minus the hole): Step \(n\) is a square of side \(n + 2\) with a square hole of side \(n\).
Step 1: \(3^2 - 1^2 = 8\).
Step 2: \(4^2 - 2^2 = 12\).
Step 3: \(5^2 - 3^2 = 16\).
Step \(n\): \((n + 2)^2 - n^2 = n^2 + 4n + 4 - n^2 = 4n + 4\).
Method 2 (four equal strips): go round the ring in four strips, each \((n + 1)\) tiles long.
Number of tiles \(= 4(n + 1) = 4n + 4\).
Method 3 (corners and sides): 4 corner tiles and 4 sides with \(n\) tiles each.
Number of tiles \(= 4 + 4n\).
Step 4: \(6^2 - 4^2 = 36 - 16 = 20\) tiles, and \(4 \times 4 + 4 = 20\).
Step 10: \(12^2 - 10^2 = 144 - 100 = 44\) tiles, and \(4 \times 10 + 4 = 44\).
?In-text Question — Find the area of the (interior) shaded region in the figure below. All four rectangles have the same dimensions. Tadang’s method gives the area (m + n)² – 4mn. Yusuf’s method gives the area (n – m)². By expanding both expressions, check that (m + n)² – 4mn = (n – m)².
Tadang’s expression: \((m + n)^2 - 4mn\)
\(= m^2 + 2mn + n^2 - 4mn\) (using \((a + b)^2 = a^2 + 2ab + b^2\))
\(= m^2 - 2mn + n^2\)
Yusuf’s expression: \((n - m)^2\)
\(= n^2 - 2nm + m^2\) (using \((a - b)^2 = a^2 - 2ab + b^2\))
\(= m^2 - 2mn + n^2\)
Both expansions give \(m^2 - 2mn + n^2\), so they are equal.
?In-text Question — Find out the area of the region with slanting lines in the figure. All three rectangles have the same dimensions (Fig. 1). Anusha’s expression is x² – xy, Vaishnavi’s is x (x + 2y) – 3xy and Aditya’s is x (x – y). By expanding the expressions, verify that all three expressions are equivalent. If x = 8 and y = 3, find the area of the shaded region.
Anusha: \(x^2 - xy\).
Vaishnavi: \(x(x + 2y) - 3xy\)
\(= x^2 + 2xy - 3xy\)
\(= x^2 - xy\)
Aditya: \(x(x - y)\)
\(= x^2 - xy\)
All three expansions give \(x^2 - xy\), so the expressions are equivalent.
For \(x = 8, y = 3\): \(x^2 - xy = 64 - 24 = 40\).
Check with Aditya’s form: \(x(x - y) = 8 \times 5 = 40\).
MTMath Talk — Write an expression for the area of the dashed region in the figure below. Use more than one method to arrive at the answer. Substitute p = 6, r = 3.5, and s = 9, and calculate the area.
The dashed region is a rectangle.
Its length is \(s - r\) and its height is \(p - r\).
Method 1 (area of the rectangle):
Area \(= (s - r)(p - r)\)
\(= s(p - r) - r(p - r)\)
\(= sp - sr - rp + r^2\)
Method 2 (whole rectangle minus the pink L-shape):
The whole rectangle is \(s \times p\).
The pink L-shape is a bottom strip \(s \times r\) and a right strip \((p - r) \times r\).
Area of dashed region \(= sp - sr - (p - r)r\)
\(= sp - sr - pr + r^2\)
Both methods give \(ps - pr - sr + r^2\).
Substitute \(p = 6, r = 3.5, s = 9\):
\((s - r)(p - r) = (9 - 3.5)(6 - 3.5) = 5.5 \times 2.5\)
\(= 13.75\)
Check: \(54 - 21 - 31.5 + 12.25 = 13.75\).
Figure it Out — Set 3 (Pages 154–156)
Eleven questions from page 154, page 155 and page 156 of the textbook.
1Compute these products using the suggested identity.
(i) 46² using Identity 1A for (a + b)²
(ii) 397 × 403 using Identity 1C for (a + b) (a – b)
(iii) 91² using Identity 1B for (a – b)²
(iv) 43 × 45 using Identity 1C for (a + b) (a – b)
(i) \(46 = 40 + 6\)
\(46^2 = 40^2 + 6^2 + 2 \times 40 \times 6\)
\(= 1600 + 36 + 480\)
\(= 2116\)
(ii) \(397 \times 403 = (400 - 3)(400 + 3)\)
\(= 400^2 - 3^2\)
\(= 160000 - 9\)
\(= 159991\)
(iii) \(91 = 100 - 9\)
\(91^2 = 100^2 + 9^2 - 2 \times 100 \times 9\)
\(= 10000 + 81 - 1800\)
\(= 8281\)
(iv) \(43 \times 45 = (44 - 1)(44 + 1)\)
\(= 44^2 - 1^2\)
\(= 1936 - 1\)
\(= 1935\)
(ii) 159991
(iii) 8281
(iv) 1935
2Use either a suitable identity or the distributive property to find each of the following products.
(i) (p – 1) (p + 11)
(ii) (3a – 9b) (3a + 9b)
(iii) –(2y + 5) (3y + 4)
(iv) (6x + 5y)²
(v) (2x – 1/2)²
(vi) (7p) × (3r) × (p + 2)
(i) \((p - 1)(p + 11)\)
\(= p(p + 11) - 1(p + 11)\)
\(= p^2 + 11p - p - 11\)
\(= p^2 + 10p - 11\)
(ii) \((3a - 9b)(3a + 9b)\)
\(= (3a)^2 - (9b)^2\) (Identity 1C)
\(= 9a^2 - 81b^2\)
(iii) \(-(2y + 5)(3y + 4)\)
\((2y + 5)(3y + 4) = 2y(3y + 4) + 5(3y + 4)\)
\(= 6y^2 + 8y + 15y + 20\)
\(= 6y^2 + 23y + 20\)
With the minus sign: \(-6y^2 - 23y - 20\)
(iv) \((6x + 5y)^2\)
\(= (6x)^2 + (5y)^2 + 2 \times 6x \times 5y\) (Identity 1A)
\(= 36x^2 + 25y^2 + 60xy\)
\(= 36x^2 + 60xy + 25y^2\)
(v) \(\left(2x - \frac{1}{2}\right)^2\)
\(= (2x)^2 + \left(\frac{1}{2}\right)^2 - 2 \times 2x \times \frac{1}{2}\) (Identity 1B)
\(= 4x^2 + \frac{1}{4} - 2x\)
\(= 4x^2 - 2x + \frac{1}{4}\)
(vi) \((7p) \times (3r) \times (p + 2)\)
\(= 21pr(p + 2)\)
\(= 21p^2r + 42pr\)
(ii) 9a² − 81b²
(iii) −6y² − 23y − 20
(iv) 36x² + 60xy + 25y²
(v) 4x² − 2x + 1/4
(vi) 21p²r + 42pr
3For each statement identify the appropriate algebraic expression(s).
(i) Two more than a square number. 2 + s, (s + 2)², s² + 2, s² + 4, 2s², 22s
(ii) The sum of the squares of two consecutive numbers. m² + n², (m + n)², m² + 1, m² + (m + 1)², m² + (m – 1)², (m + (m + 1))², (2m)² + (2m + 1)²
(i) A square number is \(s^2\).
Two more than it is \(s^2 + 2\).
\(2 + s\) is 2 more than \(s\), not than a square.
\((s + 2)^2\) is the square of \(s + 2\).
\(s^2 + 4\) is 4 more than a square.
\(2s^2\) is twice a square.
\(22s\) is 22 times \(s\).
Only \(s^2 + 2\) matches.
(ii) We need the squares of two numbers that are consecutive, added together.
\(m^2 + (m + 1)^2\): the numbers \(m\) and \(m + 1\) are consecutive.
Matches.
\(m^2 + (m - 1)^2\): the numbers \(m\) and \(m - 1\) are consecutive.
Matches.
\((2m)^2 + (2m + 1)^2\): the numbers \(2m\) and \(2m + 1\) are consecutive.
Matches.
\(m^2 + n^2\): \(m\) and \(n\) can be any numbers.
Does not match.
\((m + n)^2\) and \((m + (m + 1))^2\) are squares of sums, not sums of squares.
Do not match.
\(m^2 + 1\): the second number is 1, which is not consecutive to \(m\) in general.
Does not match.
(ii) m² + (m + 1)², m² + (m − 1)² and (2m)² + (2m + 1)²
4Math Talk — Consider any 2 by 2 square of numbers in a calendar, as shown in the figure. Find products of numbers lying along each diagonal — 4 × 12 = 48, 5 × 11 = 55. Do this for the other 2 by 2 squares. What do you observe about the diagonal products? Explain why this happens. Hint: Label the numbers in each 2 by 2 square as a, (a + 1) in the first row and a + 7, (a + 8) in the second row. (February calendar: Su M Tu W Th F Sa; 1 falls on Sa; 2 to 8, 9 to 15, 16 to 22, 23 to 28 in the next rows.)February Su M Tu W Th F Sa 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28
| February | ||||||
|---|---|---|---|---|---|---|
| Su | M | Tu | W | Th | F | Sa |
| 1 | ||||||
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 9 | 10 | 11 | 12 | 13 | 14 | 15 |
| 16 | 17 | 18 | 19 | 20 | 21 | 22 |
| 23 | 24 | 25 | 26 | 27 | 28 |
Examples:
Square 4, 5, 11, 12: \(4 \times 12 = 48\) and \(5 \times 11 = 55\).
Difference \(= 7\).
Square 3, 4, 10, 11: \(3 \times 11 = 33\) and \(4 \times 10 = 40\).
Difference \(= 7\).
Square 10, 11, 17, 18: \(10 \times 18 = 180\) and \(11 \times 17 = 187\).
Difference \(= 7\).
Observation: the two diagonal products always differ by 7.
The product along the diagonal going down to the left, \((a + 1)(a + 7)\), is bigger by 7.
Why: label the square \(a,\ a + 1\) in the top row and \(a + 7,\ a + 8\) in the bottom row.
The next row of a calendar is 7 more.
Diagonal 1: \(a(a + 8) = a^2 + 8a\).
Diagonal 2: \((a + 1)(a + 7) = a^2 + 7a + a + 7 = a^2 + 8a + 7\).
Difference \(= (a^2 + 8a + 7) - (a^2 + 8a) = 7\).
5Verify which of the following statements are true.
(i) (k + 1) (k + 2) – (k + 3) is always 2.
(ii) (2q + 1) (2q – 3) is a multiple of 4.
(iii) Squares of even numbers are multiples of 4, and squares of odd numbers are 1 more than multiples of 8.
(iv) (6n + 2)² – (4n + 3)² is 5 less than a square number.
(i) False.
\((k + 1)(k + 2) - (k + 3) = k^2 + 3k + 2 - k - 3\)
\(= k^2 + 2k - 1\)
This depends on \(k\).
For \(k = 2\): \(3 \times 4 - 5 = 7\), not 2.
(ii) False.
\((2q + 1)(2q - 3) = 4q^2 - 6q + 2q - 3\)
\(= 4q^2 - 4q - 3\)
\(= 4(q^2 - q) - 3\)
This is 3 less than a multiple of 4 (also it is odd), so it is never a multiple of 4.
Example: \(q = 2\): \(5 \times 1 = 5\), not a multiple of 4.
(iii) True.
Even number \(2m\): \((2m)^2 = 4m^2\).
This is a multiple of 4.
Odd number \(2m + 1\): \((2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1\).
\(m(m + 1)\) is a product of two consecutive numbers, so it is even.
Write it as \(2t\).
\(4m(m + 1) = 8t\).
So \((2m + 1)^2 = 8t + 1\), which is 1 more than a multiple of 8.
Example: \(7^2 = 49 = 8 \times 6 + 1\).
(iv) False.
\((6n + 2)^2 - (4n + 3)^2 = (36n^2 + 24n + 4) - (16n^2 + 24n + 9)\)
\(= 20n^2 - 5\)
For this to be 5 less than a square, \(20n^2\) would have to be a square number.
For \(n = 1\): \(20n^2 = 20\), which is not a square.
(For \(n = 2\): 80, not a square.)
So the statement is not true in general.
(ii) False
(iii) True
(iv) False
6A number leaves a remainder of 3 when divided by 7, and another number leaves a remainder of 5 when divided by 7. What is the remainder when their sum, difference, and product are divided by 7?
Let the first number be \(7a + 3\) and the second be \(7b + 5\).
Sum: \((7a + 3) + (7b + 5) = 7a + 7b + 8\)
\(= 7(a + b + 1) + 1\)
The remainder is 1.
Difference: \((7b + 5) - (7a + 3) = 7b - 7a + 2\)
\(= 7(b - a) + 2\)
The remainder is 2.
(The second number minus the first.)
If we subtract the other way, \((7a + 3) - (7b + 5) = 7(a - b - 1) + 5\), and the remainder is 5.
Product: \((7a + 3)(7b + 5) = 49ab + 35a + 21b + 15\)
\(= 7(7ab + 5a + 3b + 2) + 1\) (since \(15 = 14 + 1\))
The remainder is 1.
7Choose three consecutive numbers, square the middle one, and subtract the product of the other two. Repeat the same with other sets of numbers. What pattern do you notice? How do we write this as an algebraic equation? Expand both sides of the equation to check that it is a true identity.
Examples:
\(4, 5, 6\): \(5^2 - 4 \times 6 = 25 - 24 = 1\).
\(9, 10, 11\): \(10^2 - 9 \times 11 = 100 - 99 = 1\).
\(19, 20, 21\): \(20^2 - 19 \times 21 = 400 - 399 = 1\).
Pattern: the result is always 1.
Algebraic equation: let the numbers be \(n - 1, n, n + 1\).
The equation is \(n^2 - (n - 1)(n + 1) = 1\).
Left side: \(n^2 - (n - 1)(n + 1)\)
\(= n^2 - (n^2 + n - n - 1)\)
\(= n^2 - n^2 + 1\)
\(= 1\)
Right side: 1.
Both sides are equal for every \(n\), so it is a true identity.
8What is the algebraic expression describing the following steps — add any two numbers. Multiply this by half of the sum of the two numbers? Prove that this result will be half of the square of the sum of the two numbers.
Let the two numbers be \(a\) and \(b\).
Add them: \(a + b\).
Half of the sum: \(\frac{1}{2}(a + b)\).
Multiply: \((a + b) \times \frac{1}{2}(a + b)\)
\(= \frac{1}{2}(a + b)(a + b)\)
\(= \frac{1}{2}(a + b)^2\)
This is half of the square of the sum of the two numbers.
Example: \(a = 3, b = 5\): \(8 \times 4 = 32\) and \(\frac{1}{2} \times 8^2 = 32\).
9Which is larger? Find out without fully computing the product.
(i) 14 × 26 or 16 × 24
(ii) 25 × 75 or 26 × 74
(i) Both pairs add up to 40.
Write the numbers around 20.
\(14 \times 26 = (20 - 6)(20 + 6) = 20^2 - 6^2\)
\(16 \times 24 = (20 - 4)(20 + 4) = 20^2 - 4^2\)
\(6^2\) is bigger than \(4^2\), so \(20^2 - 6^2\) is smaller than \(20^2 - 4^2\).
So \(16 \times 24\) is larger.
(ii) Both pairs add up to 100.
Write the numbers around 50.
\(25 \times 75 = (50 - 25)(50 + 25) = 50^2 - 25^2\)
\(26 \times 74 = (50 - 24)(50 + 24) = 50^2 - 24^2\)
\(25^2\) is bigger than \(24^2\), so \(50^2 - 25^2\) is smaller than \(50^2 - 24^2\).
So \(26 \times 74\) is larger.
(ii) 26 × 74 is larger.
10A tiny park is coming up in Dhauli. The plan is shown in the figure. The two square plots, each of area g² sq. ft., will have a green cover. All the remaining area is a walking path w ft. wide that needs to be tiled. Write an expression for the area that needs to be tiled.
From the figure, the gap between the two squares is \(2w\) (this is shown in the figure by the arrow between them), and the other gaps are \(w\).
Each square plot has side \(g\), because its area is \(g^2\).
Length of the park \(= w + g + 2w + g + w = 2g + 4w\).
Breadth of the park \(= w + g + w = g + 2w\).
Area of the park \(= (2g + 4w)(g + 2w)\)
\(= 2g^2 + 4gw + 4gw + 8w^2\)
\(= 2g^2 + 8gw + 8w^2\)
Area of the two squares \(= 2g^2\).
Area to be tiled \(= 2g^2 + 8gw + 8w^2 - 2g^2\)
\(= 8gw + 8w^2\)
\(= 8w(g + w)\)
11For each pattern shown below,
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.
(i) Draw the next figure in the sequence.
(ii) How many basic units are there in Step 10?
(iii) Write an expression to describe the number of basic units in Step y.
Pattern (a): the yellow figures
(i) Each step has a block of \(y\) rows and \((y + 2)\) columns in the middle.
A long arm of \((y + 2)\) cells goes up on the right and an arm of \((y + 2)\) cells goes down on the left.
(ii) Counting the cells: Step 1 has \(3 \times 3 = 9\), Step 2 has \(4 \times 4 = 16\), Step 3 has \(5 \times 5 = 25\).
(iii) Step \(y\): the block has \(y(y + 2)\) cells and the two arms have \(2(y + 2)\) cells.
Total \(= y(y + 2) + 2(y + 2) = (y + 2)(y + 2) = (y + 2)^2\).
Step 4 has \(6^2 = 36\) cells.
Step 10 has \((10 + 2)^2 = 144\) cells.
Pattern (b): the blue figures
(i) Each step is a square of side \((y + 1)\) with \(y\) more units added in a row below.
(ii) Step 1 has \(2^2 + 1 = 5\), Step 2 has \(3^2 + 2 = 11\), Step 3 has \(4^2 + 3 = 19\), Step 4 has \(5^2 + 4 = 29\).
(iii) Step \(y\) has \((y + 1)^2 + y\) units.
Step 10 has \(11^2 + 10 = 121 + 10 = 131\) units.
Note: in the textbook figure for pattern (a), Step 3 is drawn with 6 cells in each long arm, which gives 27 cells.
Steps 1 and 2 have arms of 3 and 4 cells, so Step 3 should have arms of 5 cells (25 cells), which fits \((y + 2)^2\) and the textbook’s own answer.
The figure above is drawn with 5 cells.
Coin Conjoin
?In-text Question — Arrange 10 coins in a triangle. The task is to turn the triangle upside down by moving one coin at a time. How many moves are needed? What is the minimum number of moves? A triangle of 3 coins can be inverted (turned upside down) with a single move, and a triangle of 6 coins can be inverted by moving 2 coins. The 10-coin triangle can be flipped with just 3 moves; did you figure out how? Find out the minimum possible moves needed to flip the next bigger triangle having 15 coins. Try the same for bigger triangular numbers. Is there a simple way to calculate the minimum number of coin moves needed for any such triangular arrangement?
The idea: the flipped triangle can be placed anywhere.
We keep as many coins as possible in place, and move only the others.
10 coins (4 rows): 7 coins stay where they are (yellow).
Only 3 coins move (orange): the top coin goes to the bottom, and the two bottom corner coins go to the two ends of the second row.
15 coins (5 rows): 10 coins stay and 5 coins move.
Why not fewer: trying every position of the flipped triangle, the most coins that can stay in place is 10 for 15 coins, so at least 5 must move.
Bigger triangles:
| Coins | 3 | 6 | 10 | 15 | 21 | 28 | 36 | 45 | 55 |
|---|---|---|---|---|---|---|---|---|---|
| Minimum moves | 1 | 2 | 3 | 5 | 7 | 9 | 12 | 15 | 18 |
Pattern: the minimum number of moves is the number of coins divided by 3, with the remainder ignored.
(We checked every position of the flipped triangle for each size in the table.)
For example, \(15 \div 3 = 5\), \(21 \div 3 = 7\), \(28 \div 3 = 9\) (remainder ignored), \(55 \div 3 = 18\) (remainder ignored).
Continue with Chapter 7
Move on to Proportional Reasoning next.