Verify the identity ${(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ for $a = 2, b = 4$

Open in App

Solution

taking $a = 2, b = 4$

$(a + b)^{2} = (2 + 4)^{2} = 6^{2} = 36$

$a^{2} + b^{2} + 2 a b = 2^{2} + 4^{2} + 2 \times 2 \times 4 = 4 + 16 + 16 = 36$

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

Suggest Corrections

Similar questions

{(a + b)}^{2} \equiv a^{2} + 2 a b + b^{2}

a = 3

b = 2

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

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

{(a - b)}^{2}

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

a = 2

b = 4

c = 6

a = 5

b = - 2

c = 3

a = - 2

b = 3

c = - 4

Join BYJU'S Learning Program