Alg2.1.10

Find the value of $\sqrt{3+\sqrt{5}}\,$

Let $\sqrt{3+\sqrt{5}}=\sqrt{x}+\sqrt{y}\,$

Squaring both sides we get

$3+\sqrt{5}=x+y+2\sqrt{xy}\,$

Comparing the values both sides

$x+y=3, 2\sqrt{xy}=\sqrt{5}\,$

$x+y=16, 4xy=5\,$

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

Substituting the values in the above,

$(x-y)^2=(3)^2-5\,$

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

Hence

$(x-y)=2\,$

Solving

$(x-y)=2, (x+y)=3\,$

We get the values as

$x=\frac{5}{2},y=\frac{1}{2}\,$

Hence

$\sqrt{3+\sqrt{5}}=\sqrt{\frac{5}{2}}+\sqrt{\frac{1}{2}}\,$