Evaluate the following products without multiplying directly:
(i) $ 103 \times 107 $
(ii) $ 95 \times 96 $
(iii) $ 104 \times 96 $

AcademicMathematicsNCERTClass 9

To do:

We have to evaluate the given products without multiplying directly.

Solution:

We know that,

$(x + a)(x+b) = x( x+ b) + a(x+b)$

$=x^2+xb+ax+ab$

$=x^2+x(a+b)+ab$

Therefore,

(i) \( 103 \times 107=(100+3) \times (100+7) \)

Here, $x=100, a=3$ and $b=7$

This implies,

$(100+3)(100+7) = (100)^2+100(3+7)+3\times7$

$=10000+100(10)+21$

$=10000+1000+21$

$=11021$

Hence, $103 \times 107=11021$.

(ii) \( 95 \times 96=(100-5)\times(100-4) \)

Here, $x=100, a=-5$ and $b=-4$

This implies,

$(100-5)(100-4) = (100)^2+100(-5-4)+(-5)\times(-4)$

$=10000+100(-9)+20$

$=10000-900+20$

$=9120$

Hence, $95 \times 96=9120$.

(iii) \( 104 \times 96=(100+4)\times(100-4) \)

Here, $x=100, a=4$ and $b=-4$

This implies,

$(100+4)(100-4) = (100)^2+100(4-4)+4\times(-4)$

$=10000+100(0)-16$

$=10000+0-16$

$=9984$

Hence, $104 \times 96 =9984$.

raja
Updated on 10-Oct-2022 13:39:07

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