Prove the following identities :

$(i) (a + b)^{2} = a^{2} + b^{2} + 2 a b$

$(i i) (a - b)^{2} = a^{2} + b^{2} - 2 a b$ [2 MARKS]

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Solution

Proof: 1 Mark each

$(i) (a + b)^{2} = (a + b) (a + b)$

$= a^{2} + a b + b a + b^{2}$

$= a^{2} + a b + a b + b^{2} [a b = b a]$

$= a^{2} + b^{2} + 2 a b$

$\Rightarrow (a + b)^{2} = a^{2} + b^{2} + 2 a b$

$(i i) (a - b)^{2} = (a - b) (a - b)$

$= a^{2} - b a - a b + b^{2}$

$= a^{2} - a b - a b + b^{2} [a b = b a]$

$= a^{2} + b^{2} - 2 a b$

$\Rightarrow (a - b)^{2} = a^{2} + b^{2} - 2 a b$

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Similar questions

Q. Prove

{(a + b)}^{2} = a^{2} + 2 a b + b^{2}

Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

Q. Question 3

Which of the following is correct?
a)

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

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

(a - b)^{2} = a^{2} - b^{2}

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

Q. What is the expansion for the identity

{(a - b)}^{2}

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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Prove the following identities : (i) (a+b)2=a2+b2+2ab (ii) (a−b)2=a2+b2−2ab [2 MARKS]

Proof: 1 Mark each (i) (a+b)2=(a+b)(a+b) =a2+ab+ba+b2 =a2+ab+ab+b2[ab=ba] =a2+b2+2ab ⇒(a+b)2=a2+b2+2ab (ii)(a−b)2=(a−b)(a−b) =a2−ba−ab+b2 =a2−ab−ab+b2[ab=ba] =a2+b2−2ab ⇒(a−b)2=a2+b2−2ab

Prove the following identities :

$(i) (a + b)^{2} = a^{2} + b^{2} + 2 a b$

$(i i) (a - b)^{2} = a^{2} + b^{2} - 2 a b$ [2 MARKS]