How to find square root of a binomial quadratic surd

Assume \sqrt{a+ \sqrt{b} + \sqrt{c} + \sqrt{d}}=\sqrt{x} + \sqrt{y} + \sqrt{z};

Hence, a+\sqrt{b} + \sqrt{c} + \sqrt{d} = x+y+z+ 2\sqrt{xy} + 2\sqrt{yz}+ 2\sqrt{zx}

If then, 2\sqrt{xy}=\sqrt{b}, 2\sqrt{yz}=\sqrt{c}, 2\sqrt{zx}=\sqrt{d},

And, if simultaneously, the values of x, y, z thus found satisfy x+y+z=a, we shall have obtained the required root.

Example:

Find the square root of 21-4\sqrt{5}+5\sqrt{3}-4\sqrt{15}.

Solution:

Clearly, we can’t have anything like

21--4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} + \sqrt{y} +\sqrt{z}

We will have to try the following options:

21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x} - \sqrt{y} - \sqrt{z}

21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x}-\sqrt{y}+\sqrt{z}

21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=\sqrt{x}+\sqrt{y}-\sqrt{z}.

Only the last option will work as we now show:

So, once again, assume that \sqrt{21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}}=\sqrt{x}+\sqrt{y}-\sqrt{z}

Hence, 21-4\sqrt{5}+8\sqrt{3}-4\sqrt{15}=z+y+z+2\sqrt{xy}-2\sqrt{yz}+2\sqrt{zx}

Put 2\sqrt{xy}=8\sqrt{3}, 2\sqrt{xz}=4\sqrt{15}, 2\sqrt{yz}=4\sqrt{5};

by multiplication, xyz = 240; that is \sqrt{xyz}=4\sqrt{15}; so it follows that : \sqrt{x}=2\sqrt{3}, \sqrt{y}=2, \sqrt{z}=\sqrt{5}.

And, since, these values satisfy the equation x+y+z=21, the required root is 2\sqrt{3}+2-\sqrt{5}.

That is all, for now,

Regards,

Nalin Pithwa

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