What must be subtracted from $3a^2-6ab-3b^2-1$ to get $4a^2-7ab-4b^2-1$?

To do: To find the number that should be substracted from  $3a^2-6ab-3b^2-1$ to get $4a^2-7ab-4b^2-1$?

Solution:

 Let the number to be subtracted is $x$

$3 a^{2}   - 6 a b - 3 b^{2}   - 1  - x  =  4 a^{2}   - 7 a b - 4 b^{2}   + 1$

Bring x to the opposite side, 

$3 a ^2  - 6 a b - 3 b ^2  - 1  - (4 a ^2  - 7 a b - 4 b ^2  + 1)  =  x$

Multiply - inside the brackets,

$3 a^2  - 6 a b - 3 b^2  - 1 - 4 a 2  + 7 a b + 4 b^2  - 1  =  x$

$3 a^2  - 4 a^2   - 6 a b + 7 a b - 3 b^2   + 4 b^2  - 1 - 1  =  x$

-$ a^2  + a b  + b^ 2  - 2  =  x$

Rewrite,

$x = - a^ 2  + a b  + b^ 2  - 2$ 

So, $- a^ 2  + a b  + b^ 2  - 2$   is to be subtracted from $3 a^2  - 6 a b - 3 b^2  - 1$ ,  to get

$4 a^2  - 7 a b - 4 b^2  + 1$

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Updated on: 10-Oct-2022

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