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Ex.9.5 Q3 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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Question

Find the following squares by using the identities.

(i) \(\left( b-7 \right)^{2}\)

(ii) \(( xy+3z )^{2} \)

(iii) \(\begin{align} {{\left( 6{{x}^{2}}-5y \right)}^{2}} \end{align}\)

(iv) \(\begin{align} {{\left( \frac{2}{3}m+\frac{3}{2}n \right)}^{2}} \end{align}\)

(v) \(\begin{align} {{\left( 0.4p-0.5q \right)}^{2}} \end{align}\)

(vi) \( ( 2xy+5y )^{2} \)

 Video Solution
Algebraic Expressions & Identities
Ex 9.5 | Question 3

Text Solution

What is known?

Expressions

What is unknown?

Simplification

Reasoning:

\[\begin{align}{(a + b)^2} &= {a^2} + 2ab + {b^2}\\{(a - b)^2} &= {a^2} - 2ab + {b^2}\\(a + b)(a - b) &= {a^2} - {b^2}\end{align}\]

Steps:

(i) 

\[\begin{align} {{\left( b-7 \right)}^{2}}&= {\left( b \right)^2} - 2\left( b \right)\left( 7 \right) + {\left( 7 \right)^2} \\  [ \left( {a - b} \right)^2 &= {a^2}  - 2ab + {b^2} ]\\&= {b^2} - 14b + 49\end{align}\]

(ii) 

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

(iii)

\[\begin{align}{{\left( 6{{x}^{2}}-5y \right)}^{2}} &= \begin{bmatrix}\left( {6{x^2}} \right)^2 - 2\left( {6{x^2}} \right)\left( {5y} \right) \\ + \left( {5y} \right)^2 \end{bmatrix} \\ [\left( {a - b} \right)^2 &= {a^2} - 2ab + {b^2}] \\&= 36{x^4} - 60{x^2}y + 25{y^2}\end{align}\]

(iv)

\[\begin{align}{{\left( \frac{2}{3}m\!+\!\frac{3}{2}n \right)}^{2}} &\!=\!\!\! \begin{bmatrix}\!\left( {\frac{2}{3}m} \right)^2 \!+\! 2\left( {\frac{{\not\!2}}{{\not\!3}}m} \right)\!\left( {\frac{{\not\!3}}{{\not\!2}}n} \right) \!\!\\ + \left( {\frac{3}{2}n} \right)^2 \end{bmatrix} \\ [ \left( {a + b} \right)^2& = {a^2} + 2ab + {b^2} ]\\& = \frac{4}{9}{m^2} + 2mn + \frac{9}{4}{n^2}\end{align}\]

(v)

\[\begin{align}{{\left( 0.4p\!-\!0.5q \right)}^{2}}&\!=\!\!\! \begin{bmatrix}\! {\left( {0.4p} \right)^2}\! - \!2\left( {0.4p} \right) \left( {0.5q} \right) \\ + \left( {0.5q} \right)^2 \!\end{bmatrix} \\ [\left( {a - b} \right)^2 &= {a^2} - 2ab + {b^2}] \\&= 0.16{p^2} \!-\! 0.4pq\! + \!0.25{q^2}\end{align}\]

(vi)

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