Use a suitable identity to get each of the following products.
(i) \[(x + 3)(x + 3)\]
(ii) $(2y + 5)(2y + 5)$
(iii) $(2a - 7)(2a - 7)$
(iv) $(3a - \dfrac{1}{2})(3a - \dfrac{1}{2})$
(v) \[(1.1m - 0.4)(1.1m + 0.4)\]
(vi) \[({a^2} + {b^2})( - {a^2} + {b^2})\]
(vii) \[(6x - 7)(6x + 7)\]
(viii) \[( - a + c)( - a + c)\]
(ix) $(\dfrac{x}{2} + \dfrac{{3y}}{4})(\dfrac{x}{2} + \dfrac{{3y}}{4})$
(x) $(7a - 9b)(7a - 9b)$
Answer
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Hint: Use one of the following identities to get a polynomial of 4 terms.
\[(a + b)(a + b) = {(a + b)^2} = {a^2} + 2ab + {b^2}\] (for questions (i), (ii), and (ix))
\[(a - b)(a - b) = {(a - b)^2} = {a^2} - 2ab + {b^2}\] (for questions(iii), (iv), (vi), and (viii))
\[(a + b)(a - b) = (a - b)(a + b) = {a^2} - {b^2}\] (for questions (v) and (vii))
Combine the like terms and simplify the polynomial to get the final answer.
Complete step by step solution:
We are given 10 pairs of binomials.
We need to find the product of the binomials using suitable identities in each of these pairs.
We will be using the following identities:
\[(a + b)(a + b) = {(a + b)^2} = {a^2} + 2ab + {b^2}......(1)\]
\[(a - b)(a - b) = {(a - b)^2} = {a^2} - 2ab + {b^2}....(2)\]
\[(a + b)(a - b) = {a^2} - {b^2}....(3)\]
(i) Consider the product \[(x + 3)(x + 3)\]
We will use identity (1) to solve this.
Here$a = x$ and $b = 3$
Thus, we have \[(x + 3)(x + 3) = {x^2} + 2 \times x \times 3 + {3^2} = {x^2} + 6x + 9\]
Hence the product is\[{x^2} + 6x + 9\].
(ii) Consider the product$(2y + 5)(2y + 5)$
We will use identity (1) to solve this.
Here$a = 2y$ and $b = 5$
Thus, we have \[(2y + 5)(2y + 5) = {(2y)^2} + 2 \times 2y \times 5 + {5^2} = 4{y^2} + 20y + 25\]
Hence the product is \[4{y^2} + 20y + 25\].
(iii) Consider the product $(2a - 7)(2a - 7)$
We will use identity (2) to solve this.
Here$a = 2a$ and $b = 7$
Thus, we have \[(2a - 7)(2a - 7) = {(2a)^2} - 2 \times 2a \times 7 + {7^2} = 4{a^2} - 28a + 49\]
Hence the product is \[4{a^2} - 28a + 49\].
(iv) Consider the product $(3a - \dfrac{1}{2})(3a - \dfrac{1}{2})$
We will use identity (2) to solve this.
Here$a = 3a$ and $b = \dfrac{1}{2}$
Thus, we have \[(3a - \dfrac{1}{2})(3a - \dfrac{1}{2}) = {(3a)^2} - 2 \times 3a \times \dfrac{1}{2} + {(\dfrac{1}{2})^2} = 9{a^2} - 3a + \dfrac{1}{4}\]
Hence the product is\[9{a^2} - 3a + \dfrac{1}{4}\].
(v) Consider the product \[(1.1m - 0.4)(1.1m + 0.4)\]
We will use identity (3) to solve this.
Here$a = 1.1m$ and $b = 0.4$
Thus, we have \[(1.1m - 0.4)(1.1m + 0.4) = {(1.1m)^2} - {(0.4)^2} = 1.21{m^2} - 0.16\]
Hence the product is \[1.21{m^2} - 0.16\].
(vi) Consider the product \[({a^2} + {b^2})( - {a^2} + {b^2})\]
We can rewrite the above expression as follows: \[({b^2} + {a^2})({b^2} - {a^2})\]
We will use identity (3) to solve this.
Here$a = {b^2}$ and $b = {a^2}$
Thus, we have \[({b^2} + {a^2})({b^2} - {a^2}) = {({b^2})^2} - {({a^2})^2} = {b^4} - {a^4}\].
Hence the product is \[{b^4} - {a^4}\].
(vii) Consider the product \[(6x - 7)(6x + 7)\]
We will use identity (3) to solve this.
Here$a = 6x$ and $b = 7$
Thus, we have \[(6x - 7)(6x + 7) = {(6x)^2} - {(7)^2} = 36{x^2} - 49\].
Hence the product is\[36{x^2} - 49\].
(viii) Consider the product \[( - a + c)( - a + c)\]
We can rewrite the above expression as follows: \[(c - a)(c - a)\]
We will use identity (2) to solve this.
Here$a = c$ and $b = a$
Thus, we have \[(c - a)(c - a) = {c^2} - 2 \times c \times a + {a^2} = {c^2} - 2ac + {a^2}\]
Hence the product is \[{c^2} - 2ac + {a^2}\].
(ix) Consider the product $(\dfrac{x}{2} + \dfrac{{3y}}{4})(\dfrac{x}{2} + \dfrac{{3y}}{4})$
We will use identity (1) to solve this.
Here$a = \dfrac{x}{2}$ and $b = \dfrac{{3y}}{4}$
Thus, we have \[(\dfrac{x}{2} + \dfrac{{3y}}{4})(\dfrac{x}{2} + \dfrac{{3y}}{4}) = {(\dfrac{x}{2})^2} + 2 \times \dfrac{x}{2} \times \dfrac{{3y}}{4} + {(\dfrac{{3y}}{4})^2} = \dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}\]
Hence the product is \[\dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}\].
(x) Consider the product $(7a - 9b)(7a - 9b)$
We will use identity (2) to solve this.
Here$a = 7a$ and $b = 9b$
Thus, we have \[(7a - 9b)(7a - 9b) = {(7a)^2} + 2 \times 7a \times 9b + {(9b)^2} = 49{a^2} + 126ab + 81{b^2}\]
Hence the product is \[49{a^2} + 126ab + 81{b^2}\].
Note: When solving these type of problems please keep in mind to take care if the signs and powers associated with each of the given expressions so that you don't go wrong in the final answer
\[(a + b)(a + b) = {(a + b)^2} = {a^2} + 2ab + {b^2}\] (for questions (i), (ii), and (ix))
\[(a - b)(a - b) = {(a - b)^2} = {a^2} - 2ab + {b^2}\] (for questions(iii), (iv), (vi), and (viii))
\[(a + b)(a - b) = (a - b)(a + b) = {a^2} - {b^2}\] (for questions (v) and (vii))
Combine the like terms and simplify the polynomial to get the final answer.
Complete step by step solution:
We are given 10 pairs of binomials.
We need to find the product of the binomials using suitable identities in each of these pairs.
We will be using the following identities:
\[(a + b)(a + b) = {(a + b)^2} = {a^2} + 2ab + {b^2}......(1)\]
\[(a - b)(a - b) = {(a - b)^2} = {a^2} - 2ab + {b^2}....(2)\]
\[(a + b)(a - b) = {a^2} - {b^2}....(3)\]
(i) Consider the product \[(x + 3)(x + 3)\]
We will use identity (1) to solve this.
Here$a = x$ and $b = 3$
Thus, we have \[(x + 3)(x + 3) = {x^2} + 2 \times x \times 3 + {3^2} = {x^2} + 6x + 9\]
Hence the product is\[{x^2} + 6x + 9\].
(ii) Consider the product$(2y + 5)(2y + 5)$
We will use identity (1) to solve this.
Here$a = 2y$ and $b = 5$
Thus, we have \[(2y + 5)(2y + 5) = {(2y)^2} + 2 \times 2y \times 5 + {5^2} = 4{y^2} + 20y + 25\]
Hence the product is \[4{y^2} + 20y + 25\].
(iii) Consider the product $(2a - 7)(2a - 7)$
We will use identity (2) to solve this.
Here$a = 2a$ and $b = 7$
Thus, we have \[(2a - 7)(2a - 7) = {(2a)^2} - 2 \times 2a \times 7 + {7^2} = 4{a^2} - 28a + 49\]
Hence the product is \[4{a^2} - 28a + 49\].
(iv) Consider the product $(3a - \dfrac{1}{2})(3a - \dfrac{1}{2})$
We will use identity (2) to solve this.
Here$a = 3a$ and $b = \dfrac{1}{2}$
Thus, we have \[(3a - \dfrac{1}{2})(3a - \dfrac{1}{2}) = {(3a)^2} - 2 \times 3a \times \dfrac{1}{2} + {(\dfrac{1}{2})^2} = 9{a^2} - 3a + \dfrac{1}{4}\]
Hence the product is\[9{a^2} - 3a + \dfrac{1}{4}\].
(v) Consider the product \[(1.1m - 0.4)(1.1m + 0.4)\]
We will use identity (3) to solve this.
Here$a = 1.1m$ and $b = 0.4$
Thus, we have \[(1.1m - 0.4)(1.1m + 0.4) = {(1.1m)^2} - {(0.4)^2} = 1.21{m^2} - 0.16\]
Hence the product is \[1.21{m^2} - 0.16\].
(vi) Consider the product \[({a^2} + {b^2})( - {a^2} + {b^2})\]
We can rewrite the above expression as follows: \[({b^2} + {a^2})({b^2} - {a^2})\]
We will use identity (3) to solve this.
Here$a = {b^2}$ and $b = {a^2}$
Thus, we have \[({b^2} + {a^2})({b^2} - {a^2}) = {({b^2})^2} - {({a^2})^2} = {b^4} - {a^4}\].
Hence the product is \[{b^4} - {a^4}\].
(vii) Consider the product \[(6x - 7)(6x + 7)\]
We will use identity (3) to solve this.
Here$a = 6x$ and $b = 7$
Thus, we have \[(6x - 7)(6x + 7) = {(6x)^2} - {(7)^2} = 36{x^2} - 49\].
Hence the product is\[36{x^2} - 49\].
(viii) Consider the product \[( - a + c)( - a + c)\]
We can rewrite the above expression as follows: \[(c - a)(c - a)\]
We will use identity (2) to solve this.
Here$a = c$ and $b = a$
Thus, we have \[(c - a)(c - a) = {c^2} - 2 \times c \times a + {a^2} = {c^2} - 2ac + {a^2}\]
Hence the product is \[{c^2} - 2ac + {a^2}\].
(ix) Consider the product $(\dfrac{x}{2} + \dfrac{{3y}}{4})(\dfrac{x}{2} + \dfrac{{3y}}{4})$
We will use identity (1) to solve this.
Here$a = \dfrac{x}{2}$ and $b = \dfrac{{3y}}{4}$
Thus, we have \[(\dfrac{x}{2} + \dfrac{{3y}}{4})(\dfrac{x}{2} + \dfrac{{3y}}{4}) = {(\dfrac{x}{2})^2} + 2 \times \dfrac{x}{2} \times \dfrac{{3y}}{4} + {(\dfrac{{3y}}{4})^2} = \dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}\]
Hence the product is \[\dfrac{{{x^2}}}{4} + \dfrac{{3xy}}{4} + \dfrac{{9{y^2}}}{{16}}\].
(x) Consider the product $(7a - 9b)(7a - 9b)$
We will use identity (2) to solve this.
Here$a = 7a$ and $b = 9b$
Thus, we have \[(7a - 9b)(7a - 9b) = {(7a)^2} + 2 \times 7a \times 9b + {(9b)^2} = 49{a^2} + 126ab + 81{b^2}\]
Hence the product is \[49{a^2} + 126ab + 81{b^2}\].
Note: When solving these type of problems please keep in mind to take care if the signs and powers associated with each of the given expressions so that you don't go wrong in the final answer
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