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Algebraic Identities

iii Using x+a...

Question

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

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Solution

$103 \times 98 = (100 + 3) \times (100 - 2)$
$= (100)^{2} + (3 - 2) \times 100 + 3 \times (- 2)$
$[U s i n g i d e n t i t y (x + a) (x + b) = x^{2} + (a + b) x + a b]$
$= 10000 + (3 - 2) \times 100 - 6$
$= 10000 + 100 - 6 = 10094$

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Similar questions

Q. Question 8(i)
Using

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

103 \times 104

Q. Question 8(i)
Using

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

103 \times 104

Q. Question 8(ii)
Using

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

5.1 \times 5.2

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

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

103 \times 98

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Algebraic Identities

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