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

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Question

Show that

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

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

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

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

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

Text Solution

What is known?

LHS and RHS expression

What is unknown?

Verification of LHS \(=\) RHS

Steps:

{i} L.H.S

\[\begin{align}&= {{\left( {3x+ 7} \right)}^2} - 84x\\&=  {{\left( {3x} \right)}^2}+ 2\left( {3x} \right)\left( 7 \right) + {{\left( 7 \right)}^2}- 84x  \\&=  9{x^2}+ 42x  + 49- 84x \\&= 9{x^2}- 42x+ 49\end{align}\]

R.H.S

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

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


R.H.S 
\[\begin{align}&= {{\left( {9p + 5q} \right)}^2}\\&= {{\left( {9p} \right)}^2} \!\! + \! 2\left( {9p} \right) \!\left( {5q} \right) + \left( {5q} \right)^2  \\&= \! 81{p^2} + \!90pq +\! 25{q^2}\\ \rm{ L.H.S} &=\rm  R.H.S\end{align}\]