Find the square root of 6+sqrt{35} in the for | vedclass

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(C) To find the square root of $6+\sqrt{35}$,we multiply and divide by $2$ to make the expression a perfect square:$\sqrt{6+\sqrt{35}} = \sqrt{\frac{2

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$\frac{\sqrt{14}+\sqrt{10}}{2}$

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(C) To find the square root of $6+\sqrt{35}$,we multiply and divide by $2$ to make the expression a perfect square:$\sqrt{6+\sqrt{35}} = \sqrt{\frac{2(6+\sqrt{35})}{2}} = \sqrt{\frac{12+2\sqrt{35}}{2}}$Now,we need two numbers whose sum is $12$ and product is $35$. These numbers are $7$ and $5$.So,$12+2\sqrt{35} = 7+5+2\sqrt{7 	imes 5} = (\sqrt{7})^2 + (\sqrt{5})^2 + 2\sqrt{7}\sqrt{5} = (\sqrt{7}+\sqrt{5})^2$.Therefore,$\sqrt{\frac{12+2\sqrt{35}}{2}} = \frac{\sqrt{7}+\sqrt{5}}{\sqrt{2}}$.To rationalize the denominator,multiply numerator and denominator by $\sqrt{2}$:$\frac{(\sqrt{7}+\sqrt{5}) 	imes \sqrt{2}}{\sqrt{2} 	imes \sqrt{2}} = \frac{\sqrt{14}+\sqrt{10}}{2}$.

Find the square root of $6+\sqrt{35}$ in the form of a binomial surd.

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