Ex.9.5 Q1 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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Question

Use a suitable identity to get each of the following products.

(i) \(\begin{align} \;\;\left( x+3 \right)\left( x+3 \right) \end{align}\)

(ii) \(\begin{align}  \;\;\left( 2y+5 \right)\left( 2y+5 \right) \end{align}\)

(iii) \(\begin{align} \;\;\left( 2a-7 \right)\left( 2a-7 \right) \end{align}\)

(iv) \(\begin{align}  \;\;\left( 3a-\frac{1}{2} \right)\left( 3a-\frac{1}{2} \right) \end{align}\)

(v)  \(\begin{align}  \;\;\left( 1.1\text{m }-0.4 \right)\left( 1.1\text{ m}+0.4 \right) \end{align}\)

(vi) \(\begin{align} \;\;\left( {{a}^{2}}+{{b}^{2}} \right)\left( -{{a}^{2}}+{{b}^{2}} \right) \end{align}\)

(vii) \(\begin{align}  \;\;\left( 6x-7 \right)\left( 6x+7 \right) \end{align}\)

(viii) \(\begin{align}  \;\;\left( -a+c \right)\left( -a+c \right)~ \end{align}\)

(ix) \(\begin{align} \;\;\text{}\left( \frac{x}{2}+\frac{3y}{4} \right)\left( \frac{x}{2}+\frac{3y}{4} \right) \end{align}\)

(x) \(\begin{align}  \;\;\left( 7a-9b \right)\left( 7a-9b \right) \end{align}\)

Text Solution

What is known?

Expressions

What is unknown?

Simplification

Reasoning:

i) By using the distributive law, we can carry out the multiplication term by term.

ii) In multiplication of polynomials with polynomials, we should always look for like terms, if any, and combine them.

Steps:

The products will be as follows.

(i)  \(\begin{align}\left( x+3 \right)\left( x+3 \right)\end{align}\)

\[\begin{align}&= {\left( {x + 3} \right)^2}\\& = {{\left( x \right)}^2} + 2\left( x \right)\left( 3 \right) + {{\left( 3 \right)}^2} \\& \quad \begin{bmatrix} \left( a + b \right)^2 = {a^2} + 2ab + {b^2} \end{bmatrix}\\& = {x^2} + 6x + 9\end{align}\]

(ii) \(\begin{align}\left( 2y+5 \right)\left( 2y+5 \right)\end{align} \)

\[\begin{align}&= {\left( {2y + 5} \right)^2}\\ &= {{\left( {2y} \right)}^2} + 2\left( {2y} \right)\left( 5 \right) + {{\left( 5 \right)}^2} \\& \quad \begin{bmatrix} \left( {a + b} \right)^2= {a^2}  + 2ab + {b^2} \end{bmatrix}\\&= 4{y^2} + 20y + 25\end{align}\]

(iii) \(\begin{align} \left( {2a - 7} \right)\left( {2a - 7} \right)\end{align} \)

\[\begin{align}&= {\left( {2a - 7} \right)^2}\\&= \left( {2a} \right)^2 - 2\left( {2a} \right)\left( 7 \right)  + \left( 7 \right)^2  \\ & \quad \begin{bmatrix} \left( {a - b} \right)^2 = {a^2}  - 2ab + {b^2} \end{bmatrix} \\& = 4{a^2} - 28a + 49\end{align}\]

(iv)  \(\begin{align} \left( {3a - \frac{1}{2}} \right)\left( {3a - \frac{1}{2}} \right)\end{align} \)

\[\begin{align}& = {\left( {3a - \frac{1}{2}} \right)^2}\\& = {\left( {3a} \right)^2} - \not{\!2}\left( {3a} \right)\left( {\frac{1}{\not\!2}} \right) \! \!+\! \!{\left( {\frac{1}{2}} \right)^2} \\ & \quad [\left( {a - b} \right)^2 = {a^2}  - 2ab + {b^2} ] \\& = 9{a^2} - 3a + \frac{1}{4}\end{align}\]

(v)    \(\begin{align} ( {1.1{\rm{m }} - 0.4} ) ( {1.1{\rm{ m }} + 0.4} )\end{align} \)

\[\begin{align}&= {\left( {1.1\,\,{\rm{m}}} \right)^2} - {\left( {0.4} \right)^2}\\ & \quad [ \left( {a + b} \right)\left( {a - b} \right)  = {a^2} - {b^2}] \\&= 1.21\,\,{{\rm{m}}^2} - 0.16\end{align}\]

(vi) \(\begin{align} \left( {{a}^{2}}+{{b}^{2}} \right)\left( -{{a}^{2}}+{{b}^{2}} \right)\end{align} \)

\[\begin{align}&= \left( {{b^2} + {a^2}} \right)\left( {{b^2} - {a^2}} \right)\\&= {\left( {{b^2}} \right)^2} - {\left( {{a^2}} \right)^2} \\ & \quad[ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2} ] \\&= {b^4} - {a^4}\end{align}\]

(vii) \(\begin{align} \left( {6x - 7} \right)\left( {6x + 7} \right)\end{align} \)

\[\begin{align}&= {\left( {6x} \right)^2} - {\left( 7 \right)^2} \\ & \quad [\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}]  \\&= 36{x^2} - 49\end{align}\]

(viii)  \(\begin{align} \left( { - a + c} \right)\left( { - a + c} \right)\end{align} \)

\[\begin{align}&= {\left( { - a + c} \right)^2}\\& = {{\left( { - a} \right)}^2} + 2\left( { - a} \right)\left( c \right) + {{\left( c \right)}^2} \\ & \quad [ \left( {a + b} \right)^2 = {a^2} + 2ab + {b^2} ]\\&= {a^2} - 2ac + {c^2}\end{align}\]

(ix) \(\begin{align} \left( \frac{x}{2}+\frac{3y}{4} \right)\left( \frac{x}{2}+\frac{3y}{4} \right)\end{align} \)

\[\begin{align}& = {\left( {\frac{x}{2} + \frac{{3y}}{4}} \right)^2}\\& = 
{\left( {\frac{x}{2}} \right)^2} + {\not \!2}\left( {\frac{x}{{\not\!2}}} \right)\left( {\frac{{3y}}{4}} \right)  + {\left( {\frac{{3y}}{4}} \right)^2} \\ & \quad [\left( {a + b} \right)^2 = {a^2} + 2ab + {b^2} ] \\& = \frac{{{x^2}}}{4} + \frac{{3xy}}{4} + \frac{{9{y^2}}}{{16}}\end{align}\]

(x)  \(\begin{align} ( {7a - 9b} ) ( {7a - 9b} )\end{align} \)

\[\begin{align}&= \left( {7a - 9b} \right)^2 \\& = \left( 7a \right)^2 - 2\left( {7a} \right)\left( {9b} \right) + ( 9b )^2  \\ & \quad [\left( {a - b} \right)^2 = {a^2} - 2ab + {b^2} ] \\& { = 49{a^2} - 126ab + 81{b^2}}\end{align}\]