Verify:
$(a + b)^{2} - (a - b)^{2} = 4 a b$ , for $a = 4, b = 3$ .

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Solution

For $a = 4, b = 3$ .

Now,
$a + b = 7$ ......(1) and $a - b = 4 - 3 = 1$ .....(2).

Now, (1) $^{2} +$ (2) $^{2}$ we get,

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

$= 7^{2} - 1^{2}$

$= 49 - 1$

$= 48$ ......(3).

Now,
$4 a b$

$= 4 \times 3 \times 4$

$= 48$ .....(4).

From (3) and (4) it is varified.

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Similar questions

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

a = 2, b = 4

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

a = 3

b = 2

[{(a + b)}^{2} - 4 a b] \div (a - b)

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

(a + b)^{2} - (a - b)^{2} = 4 a b

{(a + b)}^{2} - (a - b)^{2} = 4 a b

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