Using a suitable identity, find the following.
$(3 a + 4 b) (3 a + 4 b)$
$(2 a - \frac{1}{2}) (2 a - \frac{1}{2})$

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Solution

$(i) (3 a + 4 b) (3 a + 4 b)$

$= (3 a + 4 b)^{2}$

$= (3 a)^{2} + 2 (3 a) (4 b) + (4 b)^{2}$

$= 9 a^{2} + 24 a b + 16 b^{2}$

$(i i) (2 a - \frac{1}{2}) (2 a - \frac{1}{2})$

$= {(2 a - \frac{1}{2})}^{2}$

$= (2 a)^{2} - 2 (2 a) \frac{1}{2} + {(\frac{1}{2})}^{2}$

$= 4 a^{2} - 2 a + \frac{1}{4}$

Formula:
$(x + y)^{2} = x^{2} + 2 x y + y^{2}$

$(x - y)^{2} = x^{2} - 2 x y + y^{2}$

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Similar questions

{(\frac{2}{3} a + \frac{3}{4} b)}^{2}

Use a suitable identity to get each of the following products:
$(3 a - \frac{1}{2}) (3 a - \frac{1}{2})$

(3 a - \frac{1}{2}) (3 a - \frac{1}{2})

{(2 a - 3)}^{2} - 2 (2 a - 3) (a - 1) + {(a - 1)}^{2}

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