Q. 184.2( 23 Votes )

# What must be subtracted from 3a^{2} – 6ab – 3b^{2} – 1 to get 4a^{2} – 7ab – 4ab^{2} + 1?

Answer :

Let’s suppose the required number be x,

So we have;

(3a^{2} – 6ab – 3b^{2} – 1) – x = 4a^{2} – 7a – 4b^{2} + 1

(3a^{2} – 6ab – 3b^{2} – 1) – (4a^{2} – 7a – 4b^{2} + 1) = x

So,

To get the required number we have to subtract 4a^{2} – 7a – 4b^{2} + 1 from 3a^{2} – 6ab – 3b^{2} - 1

So, the required number is - a^{2} + ab + b^{2} – 2

Rate this question :

Subtract:

5a^{2}b^{2}c^{2} from -7a^{2}b^{2}c^{2}

Take away:

(i)

(ii)

(iii)

(iv)

(v)

RD Sharma - MathematicsSubtract 3x-4y-7z from the sum of x-3y+2z and -4x+9y-11z.

RD Sharma - MathematicsSubtract:

(i) -5xy from 12xy

(ii) 2a^{2} from -7a^{2}

(iii) 2a-b from 3a-5b

(iv)

(v)

(vi)

(vii) x^{2} - xy^{2} + xy from x^{2}y + xy^{2} - xy

(viii)

RD Sharma - MathematicsIf we subtract -3x^{2}y^{2} from x^{2}y^{2}, then we get

Add:

3a(a-b-c), 2b (a-b + c)

NCERT - Mathematics ExemplarState whether the statements are true (T) or false (F).

The difference of the squares of two consecutive numbers is their sum.

NCERT - Mathematics ExemplarAdd:

xy^{2}z^{2} + 3x^{2}y^{2}z–4x^{2}yz^{2},-9x^{2}y^{2}z + 3xy^{2}z^{2} + x^{2}yz^{2}

Subtract:

2a^{2}b^{2}c^{2} + 4a^{2}b^{2}c - 5a^{2}bc^{2} from -10a^{2}b^{2}c + 4ab^{2}c^{2} + 2a^{2}bc^{2}

Add:

3a (2b + 5c), 3c (2a + 2b)

NCERT - Mathematics Exemplar