+11912 in my text book this is the question.
Example 13: Using Identity (III) , find { identity III is $$a^2 - b^2 = (a+b)(a-b)$$ }
iii. 194 x 206
(now this is the answer given in this book)
iii. 194 x 206
= (200 - 6) x (200+6)
= 200^2 - 6^2
= 40000 - 36
= 39964
now my problem is that i cant understand this answer becoz they have just applied the identity to 206 not 194 . Can anyone help me!i dont understand where 194 has gone......
+128598 Notice, rosala, that a = 200 and b = 6....so, 194 = (200 - 6) and 206 = (200 + 6)
So .....194 x 206 =
(200 - 6) x (200 + 6) = (a - b) x (a + b) =
200*200 - 6*200 + 6*200 - 6*6......notice that the two "middle terms" just "cancel" each other....so we have
2002 - 62 = a2 - b2 = 39964
And we have proved that a2 - b2 = (a -b) x (a + b)
Does that make sense??
+23246 The idea of using a + b and a - b is this:
The larger number, 206, is a + b and the smaller number, 194, is a - b.
'a' turns out to be the middle number between the larger and the smaller; in this case, a = 200 because 200 is halfways between 194 and 206.
'b' is the distance between a and the larger number and also the distance between a and the smaller number; in this case b = 6 because 200 is 6 away from both 194 and 206.
206 = 200 + 6
194 = 200 - 6
194 x 206 = (200 - 6)(200 + 6)
(a - b)(a + b)
Since (a - b)(a + b) = a² - b²
(194)(206) = (200 - 6)(200 + 6) = 200² - 6² = 40000 - 36 = 19964
Does this help?
+128598 Notice, rosala, that a = 200 and b = 6....so, 194 = (200 - 6) and 206 = (200 + 6)
So .....194 x 206 =
(200 - 6) x (200 + 6) = (a - b) x (a + b) =
200*200 - 6*200 + 6*200 - 6*6......notice that the two "middle terms" just "cancel" each other....so we have
2002 - 62 = a2 - b2 = 39964
And we have proved that a2 - b2 = (a -b) x (a + b)
Does that make sense??
