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1.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)\( (3a – \frac { 1 }{ 2 }) (3a – \frac { 1 }{ 2 })\)

(v)\( (1.1m – 0.4) (1.1m + 0.4)\)

(vi)\( (a2 + b2) (-a2 + b2)\)

(vii) \((6x – 7) (6x + 7)\)

(viii) \((-a + c) (-a + c)\)

(ix)\( (\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })\)

(x) \((7a – 9b) (7a – 9b)\)


Answer :

(i)Given: \((x + 3) (x + 3)\)
\(=(x+3)^2\)

\(=(x)^2+2\times x \times 3+(3)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=x^2+6x+9\)

(ii)Given: \((2y + 5) (2y + 5)\)
\(=(2y + 5)^2\)

\(=(2y)^2+2\times 2y \times 5+(5)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=4y^2+20y+25\)

(iii)Given: \((2a - 7) (2a - 7)\)
\(=(2a - 7)^2\)

\(=(2a)^2-2\times (2a) \times 7+(7)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=4a^2-28a+49\)

(iv)Given: \((3a - \frac{1}2) (3a - \frac{1}2)\)
\(=(3a - \frac{1}2)^2\)

\(=(3a)^2-2\times (3a) \times (\frac{1}2)+(\frac{1}2)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=9a^2-3a+\frac{1}4\)

(v)Given: \((1.1m - 0.4) (1.1m + 0.4)\)
\(=((1.1m)^2 - (0.4)^2\quad \)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=1.21a^2-0.16\)

(vi)Given: \((a^2 + b^2) (-a^2 + b^2)\)
\(=(b^2 + a^2)(b^2 -a^2) \)

\(=(b^2)^2-(a^2)^2\quad\)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=(b^4 - a^4)\)

(vii)Given: \((6x - 7) (6x + 7)\)
\(=(6x)^2 - (7)^2\quad \)[∵\((a-b)(a+b)=a^2-b^2\)]

\(=36x^2-49\)

(viii)Given: \((-a+c) (-a+c)\)
\(=[(-a)+c]^2\)

\(=(-a)^2+2\times (-a) \times c+(c)^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=a^2-2ac+c^2\)

(ix)Given:\( (\frac { x }{ 2 } + \frac { 3y }{ 4 }) (\frac { x }{ 2 } + \frac { 3y }{ 4 })\)

\(=(\frac { x }{ 2 } + \frac { 3y }{ 4 })^2\)

\(=(\frac { x }{ 2 })^2+2\times (\frac { x }{ 2 }) \times (\frac { 3y }{ 4 })+(\frac { 3y }{ 4 })^2\quad\)[∵\((a+b)^2=a^2+2ab+b^2\)]

\(=\frac { x^2 }{ 4 }+\frac { 3 }{ 4 }xy+\frac { 9y^2 }{ 16 }\)

(x)Given: \((7a - 9b) (7a - 9b)\)
\(=(7a - 9b)^2\)

\(=(7a)^2-2\times (7a) \times (9b)+(9b)^2\quad\)[∵\((a-b)^2=a^2-2ab+b^2\)]

\(=49a^2-126ab+81b^2\)