The value of sqrt{12sqrt{5} + 2sqrt{55}} is: | vedclass

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Question

(C) Let the expression be $x = \sqrt{12\sqrt{5} + 2\sqrt{55}}$.We can rewrite the expression inside the square root as $12\sqrt{5} + 2\sqrt{55} = 2\sq

Vedclass · Accepted Answer

$5^{1/4}(\sqrt{11} + 1)$

Vedclass · Answer

(C) Let the expression be $x = \sqrt{12\sqrt{5} + 2\sqrt{55}}$.We can rewrite the expression inside the square root as $12\sqrt{5} + 2\sqrt{55} = 2\sqrt{5}(6 + \sqrt{11})$.This does not immediately simplify to a perfect square of the form $(\sqrt{a} + \sqrt{b})^2$.Let us re-examine the expression: $\sqrt{12\sqrt{5} + 2\sqrt{55}} = \sqrt{2\sqrt{5} \cdot 6 + 2\sqrt{55}} = \sqrt{2\sqrt{5}(6 + \sqrt{11})}$.Alternatively,consider the form $(\sqrt{x} + \sqrt{y})^2 = x + y + 2\sqrt{xy}$.We have $12\sqrt{5} + 2\sqrt{55} = 2\sqrt{5} \cdot 6 + 2\sqrt{55} = 2\sqrt{5} \cdot \sqrt{36} + 2\sqrt{55} = 2\sqrt{180} + 2\sqrt{55}$.This is not standard. Let's try to express $12\sqrt{5} + 2\sqrt{55}$ as $(\sqrt{a} + \sqrt{b})^2 = a + b + 2\sqrt{ab}$.$a + b = 12\sqrt{5}$ and $ab = 55$.Solving $t^2 - (12\sqrt{5})t + 55 = 0$ using the quadratic formula:$t = \frac{12\sqrt{5} \pm \sqrt{(12\sqrt{5})^2 - 4(55)}}{2} = \frac{12\sqrt{5} \pm \sqrt{720 - 220}}{2} = \frac{12\sqrt{5} \pm \sqrt{500}}{2} = \frac{12\sqrt{5} \pm 10\sqrt{5}}{2}$.$t_1 = \frac{22\sqrt{5}}{2} = 11\sqrt{5}$ and $t_2 = \frac{2\sqrt{5}}{2} = \sqrt{5}$.Thus,$12\sqrt{5} + 2\sqrt{55} = (\sqrt{11\sqrt{5}} + \sqrt{\sqrt{5}})^2$.Taking the square root,we get $\sqrt{11\sqrt{5}} + \sqrt{\sqrt{5}} = \sqrt{11} \cdot 5^{1/4} + 5^{1/4} = 5^{1/4}(\sqrt{11} + 1)$.

The value of $\sqrt{12\sqrt{5} + 2\sqrt{55}}$ is:

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