Question 8iUsing x-ax-b - x2-a-bx-ab- find103- 104

Name: NCERT - Grade 08 - Mathematics - Algebraic Expressions and Identities - Q25_1
Uploaded: 2022-07-04T10:36:42+05:30
Description: 103× 104=(100+3)×(100+4) = (100)^2+(3+4)× 100 + 3× 4 [ Using identity (x+a) (x+b) =x^2+(a+b) x+ab ] =10000+7× 100+12 =10000+700+12=10712 ...

103× 104=(100+3)×(100+4) = (100)^2+(3+4)× 100 + 3× 4 [ Using identity (x+a) (x+b) =x^2+(a+b) x+ab ] =10000+7× 100+12 =10000+700+12=10712 ...

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Standard IX

Mathematics

Factors

Question 8iUs...

Question

Question 8(i)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$103 \times 104$

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Solution

$103 \times 104 = (100 + 3) \times (100 + 4)$
$= (100)^{2} + (3 + 4) \times 100 + 3 \times 4$
$[U s i n g i d e n t i t y (x + a) (x + b) = x^{2} + (a + b) x + a b]$
$= 10000 + 7 \times 100 + 12$
$= 10000 + 700 + 12 = 10712$

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Q. Question 8(i)
Using

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

, find

103 \times 104

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

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)
Using

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

, find

5.1 \times 5.2

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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Question 8(i) Using (x+a)(x+b)=x2+(a+b)x+ab, find 103×104

103×104=(100+3)×(100+4) =(100)2+(3+4)×100+3×4 [Using identity (x+a)(x+b)=x2+(a+b)x+ab] =10000+7×100+12 =10000+700+12=10712

Question 8(i)
Using $(x + a) (x + b) = x^{2} + (a + b) x + a b$ , find
$103 \times 104$

$103 \times 104 = (100 + 3) \times (100 + 4)$
$= (100)^{2} + (3 + 4) \times 100 + 3 \times 4$
$[U s i n g i d e n t i t y (x + a) (x + b) = x^{2} + (a + b) x + a b]$
$= 10000 + 7 \times 100 + 12$
$= 10000 + 700 + 12 = 10712$