Find square root of $a^{2} + 4 a b + 4 b^{2}$

a - 2 b

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a + 2 b

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a - b

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a + b

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Solution

The correct option is B $a + 2 b$
Consider the square root of the given equation $a^{2} + 4 a b + 4 b^{2}$ as shown below:

$\sqrt{a^{2} + 4 a b + 4 b^{2}} = \sqrt{(a)^{2} + (2 \times a \times 2 b) + (2 b)^{2}} = \sqrt{(a)^{2} + (2 b)^{2} + (2 \times a \times 2 b)} = \sqrt{(a + 2 b)^{2}} (∵ {(x + y)}^{2} = x^{2} + y^{2} + 2 x y) = a + 2 b$

hence, the square root of $a^{2} + 4 a b + 4 b^{2}$ is $a + 2 b$ .

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Similar questions

If a-2b = 15 and find a²+ 4b² .

a + 2 b = 5

a b = 2

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{[\frac{1}{4} a - \frac{1}{2} b + 1]}^{2}

= \frac{1}{16} a^{2} + \frac{1}{4} b^{2} + 1 - \frac{1}{4} a b - b + \frac{a}{2}

y = m x - b \sqrt{1 + m^{2}}

x^{2} + y^{2} = b^{2}

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

a > 2 b > 0

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a + 2 b = 9 and a b = 7

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