Ex.9.5 Q5 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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Question

Show that

(i)\(\begin{align}\quad{\left( {3x + 7} \right)^2} - 84x = {\left( {3x - 7} \right)^2}\end{align}\)

(ii)\(\begin{align}\quad{\left( {9p - 5q} \right)^2} + 180pq = {\left( {9p + 5q} \right)^2}\end{align}\)

(iii)\(\begin{align} \quad {{{\left( {\frac{4}{3}m - \frac{3}{4}n} \right)}^2} + 2mn = \frac{{16}}{9}{m^2} + \frac{9}{{16}}{n^2}} \end{align}\)

(iv)\(\begin{align}\quad {{{\left( {4pq{\rm{ }} + {\rm{ }}3q} \right)}^2}{\rm{ }} - {\rm{ }}{{\left( {4pq{\rm{ }} - {\rm{ }}3q} \right)}^2}{\rm{ }} = {\rm{ }}48pq} \end{align}\)

(v)\(\begin{align} \quad  {\left( {a- {}b} \right){}\left( {a+ {}b} \right)+ {}\left( {b- {}c} \right){}\left( {b+ {}c} \right)+ {}\left( {c- {}a} \right){}\left( {c+ {}a} \right)= {}0}\end{align}\)

Text Solution

What is known?

LHS and RHS expression

What is unknown?

Verification of LHS \(=\) RHS

Steps:

\(\begin{align}\left(\rm{i} \right)\quad L.H.S&= {{\left( {3x+ 7} \right)}^2} - 84x\\&= {{\left( {3x} \right)}^2}+ 2\left( {3x} \right)\left( 7 \right)+ {{\left( 7 \right)}^2}- 84x\\&= 9{x^2}+ 42x+ 49- 84x\\&= 9{x^2}- 42x+ 49\\R.H.S&= {\left( {3x- 7} \right)^2}\\&= {\left( {3x} \right)^2}- 2\left( {3x} \right)\left( 7 \right)+ {\left( 7 \right)^2}\\&= 9{x^2} - 42x+ 49\\L.H.S&= R.H.S\end{align}\)

\(\begin{align}\left( \rm{ii} \right) \quad L.H.S &= {{\left( {9p - 5q} \right)}^2} + 180pq\\&= {{\left( {9p} \right)}^2} - 2\left( {9p} \right)\left( {5q} \right) + {{\left( {5q} \right)}^2}\,\, + \,\,180pq\\&= 81{p^2} - 90pq + 25{q^2} + 180pq\\&= 81{p^2} + 90pq + 25{q^2}\\R.H.S &= {{\left( {9p + 5q} \right)}^2}\\&= {{\left( {9p} \right)}^2} + 2\left( {9p} \right)\left( {5q} \right) + {{\left( {5q} \right)}^2}\\&= 81{p^2} + 90pq + 25{q^2}\\L.H.S &= R.H.S\end{align}\)