Exercise 9.5 Algebraic Expressions and Identities- NCERT Solutions Class 8

Go back to  'Algebraic Expressions and Identities'

Question 1

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

Solution

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

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

Question 2

Use the identity

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

to find the following products.

(i) \(\begin{align} \quad \left( {x + 3} \right)\left( {x + 7} \right)\end{align}\)

(ii) \(\begin{align}\quad {\rm}\left( {4x + 5} \right)\left( {4x + 1} \right)\end{align}\)

(iiii) \(\begin{align}\quad \left( {4x - 5} \right)\left( {4x - 1} \right)\end{align}\)

(iv) \(\begin{align} \quad {\rm}\left( {4x + 5} \right)\left( {4x - 1} \right)\end{align}\)

(v) \(\begin{align}\quad {\rm}\left( {2x + 5y} \right)\left( {2x + 3y} \right)\end{align}\)

(vi)    \(\begin{align} \,\left( 2{{a}^{2}}+9 \right)\left( 2{{a}^{2}}+5 \right)\end{align}\)

(vii) \(\begin{align} \quad \left( {xyz - 4} \right)\left( {xyz - 2} \right)\end{align}\)

Solution

Video Solution

What is known?

\(\begin{align}& \left( {x + a} \right)\left( {x + b} \right) \\ &= {x^2} + \left( {a + b} \right)x + ab\end{align} \)

What is unknown?

Simplification

Steps:

The products will be as follows.

(i) \(\begin{align} \left( {x + 3} \right)\left( {x + 7} \right)\end{align}\)

\[\begin{align}&= {x^2} \! + \! \left( {3 + 7} \right)x \! +\! ( 3 )( 7 )\\& = {x^2} + 10x + 21\end{align}\]

(ii) \(\begin{align} {\rm}\left( {4x + 5} \right)\left( {4x + 1} \right)\end{align}\)

\[\begin{align}&=  {\left( {4x} \right)^2} + \left( {5 + 1} \right)\left( {4x} \right) + \left( 5 \right)\left( 1 \right) \\&= 16{x^2} + 24x + 5\end{align}\]

(iiii) \(\begin{align} \left( {4x - 5} \right)\left( {4x - 1} \right)\end{align}\)

\[\begin{align}&=\begin{bmatrix} {\left( {4x} \right)^2}\! +\! \left[ {\left( { - 5} \right) \!+\! \left( { - 1} \right)} \right] \left( {4x} \right) \\\!+\! \left( { - 5} \right)\left( { - 1} \right)\end{bmatrix} \\&= 16{x^2} - 24x + 5\end{align}\]

(iv) \(\begin{align} {\rm}\left( {4x + 5} \right)\left( {4x - 1} \right)\end{align}\)

\[\begin{align}&= \begin{bmatrix} {( {4x})^2} \!+\! \left[ {( { + 5} ) \!+ \!( { - 1} )} \right] ( {4x} ) \\\! +\! ( { + 5} )\left( { - 1} \right) \end{bmatrix}\\&= 16{x^2} + 16x - 5\end{align}\]

(v) \(\begin{align} {\rm}\left( {2x + 5y} \right)\left( {2x + 3y} \right)\end{align}\)

\[\begin{align}&=\begin{bmatrix} {\left( {2x} \right)^{2}} + \left( {5y + 3y} \right)\left( {2x} \right) \\ + \left( {5y} \right)\left( {3y} \right) \end{bmatrix} \\& = 4{x^2} + 16xy + 15{y^2}\end{align}\]

(vi) \(\begin{align} \left( 2{{a}^{2}}+9 \right)\left( 2{{a}^{2}}+5 \right)\end{align}\)

\[\begin{align}& =\begin{bmatrix} {\left( {2{a^2}} \right)^2} + \left( {9 + 5} \right)\left( {2{a^2}} \right) \\ + \left( 9 \right)\left( 5 \right) \end{bmatrix} \\ & = 4{a^4} + 28{a^2} + 45\end{align}\]

(vii) \(\begin{align} \left( {xyz - 4} \right)\left( {xyz - 2} \right)\end{align}\)

\[\begin{align}&=\begin{bmatrix} {{\left( {xyz} \right)}^2} + \left[ {\left( { - 4} \right) + \left( { - 2} \right)} \right] \\ \left( {xyz} \right) + \left( { - 4} \right)\left( { - 2} \right) \end{bmatrix} \\&= {x^2}{y^2}{z^2} - 6xyz + 8\end{align}\]

Question 3

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

Solution

Video 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)

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

Question 4

Simplify.

(i) \( ({a}^{2}-{b}^{2} )^{2} \)

(ii) \(\begin{align}{{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}\end{align}\)

(iii)\(\begin{align}{{\left( 7m-8n \right)}^{2}}+{{\left( 7m+8n \right)}^{2}}\end{align}\)

(iv)\(\begin{align}{{\left( 4m+5n \right)}^{2}}+{{\left( 5m+4n \right)}^{2}}\end{align}\)

(v)\(\begin{align}{{\left( 2.5p-1.5q \right)}^{2}}-{{\left( 1.5p-2.5q \right)}^{2}}\end{align}\)

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

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

Solution

Video 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)

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

(ii) \(\begin{align} \quad {{\left( 2x+5 \right)}^{2}}-{{\left( 2x-5 \right)}^{2}}\end{align}\)

\[\begin{align}&=\begin{Bmatrix}  {\left( {2x} \right)^2} + 2\left( {2x} \right)\left( 5 \right) + {\left( 5 \right)^2} -\\ [ \left( {2x} \right)^2 - 2\left( {2x} \right)\left( 5 \right)  + {{\left( 5 \right)}^2}] \end{Bmatrix}  \\&\qquad \left[ {{{\left( {a - b} \right)}^2} = {a^2} - 2ab + {b^2}} \right]\\&\qquad\left[ {{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right]\\& =\begin{Bmatrix} 4{x^2} + 20x + 25 \\ - \left[ {4{x^2} - 20x + 25} \right] \end{Bmatrix} \\&= \begin{Bmatrix} \not\!4{x^2} + \not\!\!20x + 25 \\ -\not\!4 {x^2} + \not\!\!20x - 25 \end{Bmatrix} \\& = 40x\end{align}\]

(iv)\(\begin{align}{{\left( 4m+5n \right)}^{2}}+{{\left( 5m+4n \right)}^{2}}\end{align}\)

\[\begin{align}&= \begin{Bmatrix} {\left( {4m} \right)^2} + 2\left( {4m} \right)\left( {5n} \right) \\ + {\left( {5n} \right)^2} + {\left( {5m} \right)^2} \\ + 2\left( {5m} \right)\left( {4n} \right) + {\left( {4n} \right)^2}\end{Bmatrix} \\ & \quad  \left[ {{{\left( {a + b} \right)}^2} = {a^2} + 2ab + {b^2}} \right]\\&=\begin{Bmatrix} 16{m^2} + 40mn \\ + 25{n^2} + 25{m^2} \\ + 40mn + 16{n^2} \end{Bmatrix}\\&= 41{m^2} + 80mn + 41{n^2}\end{align}\]

Question 5

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

Solution

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

Question 6

Using identities, evaluate.

(i) \(\begin{align}{{71^2}} \end{align}\)

(ii) \(\begin{align}{{99^2}}\end{align}\)

(iii) \(\begin{align}{{102^2}}\end{align}\)

(iv) \(\begin{align}{{998^2}}\end{align}\)

(v) \(\begin{align}{{\left( {5.2} \right)^2}}\end{align}\)

(vi) \(\begin{align}297\times 303\end{align}\)

(vii) \(\begin{align} 78\times 82\end{align}\)

(viii) \(\begin{align} {{8.9^2}}\end{align}\)

(ix) \(\begin{align} 1.05\times 9.5\end{align}\)

Solution

Video Solution

What is known?

Expressions

What is unknown?

Values of the expressions

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) \(\begin{align} \quad {{71^2}} \end{align}\)

\[\begin{align}&= {{\left( {70 + 1} \right)}^2}\\&= {{\left( {70} \right)}^2} + 2\left( {70} \right)\left( 1 \right) + {{\left( 1 \right)}^2} \\ & \quad [ \left( {a + b} \right)^2 = {a^2}  + 2ab + {b^2} ]\\&= 4900 + 140 + 1\\ &= 5041\end{align}\]

(ii) \(\begin{align} \quad {{99^2}}\end{align}\)

\[\begin{align} &={{(100-1)}^{2}} \\&=\left( 100 \right)^{2}-2\left( 100 \right)\left( 1 \right)+{{\left( 1 \right)}^{2}} \\ & \quad [ \left( a-b \right)^{2} ={{a}^{2}}  -2ab+{{b}^{2}} ]\\& =10000-200+1 \\& =9801\end{align}\]

(iii) \(\begin{align} \quad {{102^2}}\end{align}\)

\[\begin{align} &= {{\left( {100 + 2} \right)}^2}\\&= {{\left( {100} \right)}^2} + 2\left( {100} \right)\left( 2 \right) + {{\left( 2 \right)}^2} \\ & \quad  [\left( {a + b} \right)^2 = {a^2}  + 2ab + {b^2} ] \\&= 10000 + 400 + 4\\&= 10404\end{align}\]

(iv) \(\begin{align} \quad {{998^2}}\end{align}\)

\[\begin{align}&= \left( {1000 - 2} \right)^2\\&= \left( {1000} \right)^2 - 2\left( {1000} \right) \!\left( 2 \right) \!\!+ \!\!\left( 2 \right)^2 \\ & \quad [\left( {a - b} \right)^2 = {a^2}  - 2ab + {b^2}] \\&= 1000000 - 4000 + 4\\& = 996004\end{align}\]

(v) \(\begin{align} \quad {{\left( {5.2} \right)^2}}\end{align}\)

\[\begin{align}&= \left( {5.0 + 0.2}\right)^2\\&= \left( {5.0} \right)^2 \!+ 2\left( {5.0} \right) \! \left( {0.2} \right) \! + \left( \! {0.2} \! \right)^2 \\ & \quad [ \left( {a + b} \right)^2 = {a^2}  + 2ab + {b^2}]  \\&= 25 + 2 + 0.04\\&= 27.04\end{align}\]

(vi) \(\begin{align} \quad 297\times 303\end{align}\)

\[\begin{align}&= \left( {300 - 3} \right) \times \left( {300 + 3} \right)\\ &= {{\left( {300} \right)}^2} - {{\left( 3 \right)}^2} \\ & \quad [ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}]  \\&= 90000 - 9\\&= 89991\end{align}\]

(vii) \(\begin{align} \quad 78\times 82\end{align}\)

\[\begin{align}&= \left( {80 - 2} \right)\left( {80 + 2} \right)\\ &= {{\left( {80} \right)}^2} - {{\left( 2 \right)}^2} \\ & \quad[ \left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}]  \\&= 6400 - 4\\&= 6396\end{align}\]

(viii) \(\begin{align} \quad {{8.9^2}}\end{align}\)

\[\begin{align}&= \left( {9.0 - 0.1} \right)^2 \\&= \left( {9.0} \right)^2 - \!2\left( {9.0}\! \right) \!\left( {0.1} \!\right) \!+ \left( {0.1} \!\right)^2 \\ & \quad  [\left( {a - b} \right)^2 = {a^2} - 2ab + {b^2} ] \\&= 81 - 1.8 + 0.01\\&= 79.21\end{align}\]

(ix) \(\begin{align} \quad 1.05\times 9.5\end{align}\)

\[\begin{align}&= 1.05 \times 0.95 \times 10\\&= \left( {1 + 0.05} \right)\left( {1 - 0.05} \right) \times 10\\&= \left[ {{{\left( 1 \right)}^2} - {{\left( {0.05} \right)}^2}} \right] \times 10\\&= \left[ {1 - 0.0025} \right] \times 10 \\ & \quad [ \left( {a + b} \right)\left( {a - b} \right)  = {a^2} - {b^2}] \\&= 0.9975 \times 10\\&= 9.975\end{align}\]

Question 7

Using \({a^2}{\rm{ }} - {\rm{ }}{b^2} = \left( {a{\rm{ }} + {\rm{ }}b} \right){\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right),\) find

(i) \(\begin{align} \quad{{51}^2} -{{49}^2}\end{align}\)

(ii) \(\begin{align} \quad {{\left( {1.02} \right)}^2} - {{\left( {0.98} \right)}^2}\end{align}\)

(iii) \(\begin{align}\quad {{153}^2} - {{147}^2}\end{align}\)

(iv) \(\begin{align} \quad {{12.1}^2} - {{7.9}^2}\end{align}\)

Solution

Video Solution

What is known?

\({a^2}{\rm{ }} - {\rm{ }}{b^2} = \left( {a{\rm{ }} + {\rm{ }}b} \right){\rm{ }}\left( {a{\rm{ }} - {\rm{ }}b} \right),\)

What is unknown?

Results of the given expression with their corresponding values

Steps:

(i) \(\begin{align} \quad{{51}^2} -{{49}^2}\end{align}\)

\[\begin{align}&= \left( {51 + 49} \right)\left( {51 - 49} \right)\\&= \left( {100} \right)\left( 2 \right) = 200\end{align}\]

(ii) \(\begin{align} \quad {{\left( {1.02} \right)}^2} - {{\left( {0.98} \right)}^2}\end{align}\)

\[\begin{align}&= \left( {1.02 + 0.98} \right)\left( {1.02 - 0.98} \right)\\&= \left( 2 \right)\left( {0.04} \right)\\&= 0.08\end{align}\]

(iii) \(\begin{align}\quad {{153}^2} - {{147}^2}\end{align}\)

\[\begin{align}&= \left( {153 + 147} \right)\left( {153 - 147} \right)\\&= \left( {300} \right)\left( 6 \right)\\&= 1800\end{align}\]

(iv) \(\begin{align} \quad {{12.1}^2} - {{7.9}^2}\end{align}\)

\[\begin{align}&= \left( {12.1 + 7.9} \right)\left( {12.1 - 7.9} \right) \\&= \left( {20.0} \right)\left( {4.2} \right) \\&= 84\end{align}\]

Question 8

Using

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

find

(i)\(\begin{align}\quad 103\times 104\end{align}\)

(ii)\(\begin{align}\quad 5.1\times 5.2\end{align}\)

(iii)\(\begin{align}\quad 103\times 98\end{align}\)

(iv)\(\begin{align}\quad 9.7\times 9.8\end{align}\)

Solution

Video Solution

What is known?

\(\begin{align} & \left( {x + a} \right)\left( {x + b} \right) \\ &= {x^2} + \left( {a + b} \right)x + ab \end{align} \)

What is unknown?

Results of the given expression with their corresponding values

Steps:

(i)

\(\begin{align}  103 \times 104 &= \left( {100 + 3} \right)\left( {100 + 4} \right)\\&= \begin{Bmatrix} \left( {100} \right)^2 + \left( {3 + 4} \right) \\ \left( {100} \right) + \left( 3 \right)\left( 4 \right) \end{Bmatrix} \\&= 10000 + 700 + 12\\&= 10712\end{align}\)

(ii)

\(\begin{align} 5.1 \times 5.2 &= \left( {5 + 0.1} \right)\left( {5 + 0.2} \right)\\&=\begin{Bmatrix} \!\left( 5 \right)^2 + \left( {0.1 + 0.2} \right)\left( 5 \right) +\\ \left( {0.1} \right)\left( {0.2} \right) \!\! \!\end{Bmatrix}\\ &= 25 + 1.5 + 0.02\\&= 26.52\end{align}\)

(iii)

\(\begin{align}103 \times 98 &= \left( {100 + 3} \right)\left( {100 - 2} \right)\\&= \begin{Bmatrix} \left( {100} \right)^2 + \left[ {3 + \left( { - 2} \right)} \right]\left( {100} \right)+ \\ \left( 3 \right)\left( { - 2} \right) \end{Bmatrix} \\&= 10000 + 100 - 6\\&= 10094\end{align}\)

(iv)

\(\begin{align}9.7 \times 9.8 &= \left( {10 - 0.3} \right)\left( {10 - 0.2} \right)\\&=\begin{Bmatrix} \left( {10} \right)^2 + \left[ \left( { - 0.3} \right) + \left( { - 0.2} \right) \right] \\ \left( {10} \right) + \left( \! - 0.3 \!\right)\left( \!- 0.2\! \right)\!\! \end{Bmatrix} \\&= 100 + \left( { - 0.5} \right)10 + 0.06\\ &= 100 - 5 + 0.06\\&= 95 + 0.06 \\&= 95.06\end{align}\)

  
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