Alg2.1.9

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Find the value of $\sqrt{16+2\sqrt{55}}\,$

Let $\sqrt{16+2\sqrt{55}}=\sqrt{x}+\sqrt{y}\,$

Squaring both sides we get

${16+2\sqrt{55}}=x+y+2\sqrt{xy}\,$

Comparing the values both sides

$x+y=16, 2\sqrt{xy}=2\sqrt{55}\,$

$x+y=16, xy=55\,$

$(x-y)^2=(x+y)^2-4xy\,$

Substituting the values in the above,

$(x-y)^2=(16)^2-4(55)\,$

$(x-y)^2=36\,$

Hence

$(x-y)=6\,$

Solving

$(x-y)=6, (x+y)=16\,$

We get the values as

$x=11,y=5\,$

Hence

$\sqrt{16+2\sqrt{55}}=\sqrt{11}+\sqrt{5}\,$

Main Page:Algebra:Radicals

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