Q. 184.2( 23 Votes )
What must be subtracted from 3a2 – 6ab – 3b2 – 1 to get 4a2 – 7ab – 4ab2 + 1?
Answer :
Let’s suppose the required number be x,
So we have;
(3a2 – 6ab – 3b2 – 1) – x = 4a2 – 7a – 4b2 + 1
(3a2 – 6ab – 3b2 – 1) – (4a2 – 7a – 4b2 + 1) = x
So,
To get the required number we have to subtract 4a2 – 7a – 4b2 + 1 from 3a2 – 6ab – 3b2 - 1
So, the required number is - a2 + ab + b2 – 2
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