# Ex.9.5 Q8 Algebraic Expressions and Identities Solution - NCERT Maths Class 8

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## Question

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}

Video Solution
Algebraic Expressions & Identities
Ex 9.5 | Question 8

## Text 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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