## 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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