Question 8ivUsing x-ax-b - x2-a-bx-ab- find9-7-9-8

9.7× 9.8=[10 0.3]× [10 0.2] =(10)^2+(( 0.3)+( 0.2))× 10+( 0.3)× ( 0.2) [ Using identity (x+a) (x+b) =x^2 + (a+b) x + ab ] =100+{ 0.3 0.2 }× 10+0.06 =100 0.5 ...

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Standard IX

Mathematics

Factors

Question 8ivU...

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Question 8(iv)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$9.7 \times 9.8$

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Solution

$9.7 \times 9.8 = [10 - 0.3] \times [10 - 0.2]$
$= (10)^{2} + ((- 0.3) + (- 0.2)) \times 10 + (- 0.3) \times (- 0.2)$
$[U s i n g i d e n t i t y (x + a) (x + b) = x^{2} + (a + b) x + a b]$
$= 100 + {- 0.3 - 0.2} \times 10 + 0.06$
$= 100 - 0.5 \times 10 + 0.06$
$= 100 - 5 + 0.06 = 95.06$

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Question 8(iv)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$9.7 \times 9.8$

Q. Question 8(ii)
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(x + a) (x + b) = x^{2} + (a + b) x + a b

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5.1 \times 5.2

Question 8(iii)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$103 \times 98$

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(x + a) (x + b) = x^{2} + (a + b) x + a b

, find

9.7 \times 9.8

Q. Question 8(i)
Using

(x + a) (x + b) = x^{2} + (a + b) x + a b

, find

103 \times 104

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Question 8(iv) Using (x+a)(x+b)=x2+(a+b)x+ab, find 9.7×9.8

9.7×9.8=[10−0.3]×[10−0.2] =(10)2+((−0.3)+(−0.2))×10+(−0.3)×(−0.2) [Using identity (x+a)(x+b)=x2+(a+b)x+ab] =100+{−0.3−0.2}×10+0.06 =100−0.5×10+0.06 =100−5+0.06=95.06

Question 8(iv)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$9.7 \times 9.8$

$9.7 \times 9.8 = [10 - 0.3] \times [10 - 0.2]$
$= (10)^{2} + ((- 0.3) + (- 0.2)) \times 10 + (- 0.3) \times (- 0.2)$
$[U s i n g i d e n t i t y (x + a) (x + b) = x^{2} + (a + b) x + a b]$
$= 100 + {- 0.3 - 0.2} \times 10 + 0.06$
$= 100 - 0.5 \times 10 + 0.06$
$= 100 - 5 + 0.06 = 95.06$