a - b 2- a - b 2- k- where a and b are whole numbers -Then k is alwaysA- greater than aB- greater than bC- Both A and BD- None of these

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Factorisation Using Algebraic Identities

a + b 2- a - ...

Question

$(a + b)^{2} - (a - b)^{2}$ = k, where a and b are whole numbers .Then k is always

greater than a

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greater than b

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Both A and B

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None of these

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Similar questions

If $a^{2} + b^{2} + c^{2} = 14$ ;then $a b + b c + c a$ is always greater than or equal to

(- b c, b^{2} + b c, c^{2} + b c), (a^{2} + a c, - a c, c^{2} + a c)

(a^{2} + a b, b^{2} + a b, - a b)

a, b

c

\frac{a}{b} + \frac{b}{c} + \frac{c}{a}

\frac{(a + b + c) (a b + b c + c a)}{a b c} = x

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