# square root of square root binomial

Some people call the expressions of the form  $a\!+\!b\sqrt{c}$  the , especially when $c$ is an square-free integer greater than 1 (and $a$ and $b$ rational numbers).  On the high school (see e.g. division), or also polynomials containing several square root .  Taking the square root of a square root binomial is more difficult and usually results nested square roots.  However, there are some exceptions if the numbers are appropriate.  We have the

 $\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{a^{2}-% b}}{2}}$

and

 $\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{a^{2}-% b}}{2}}.$

If  $a^{2}\!-\!b$  happens to be square of a rational number, then the into expressions without nested square roots.

For example, because  $6^{2}\!-\!20=16=4^{2}$,  we obtain

 $\sqrt{6\!+\!2\sqrt{5}}=\sqrt{6\!+\!\sqrt{20}}=\sqrt{\frac{6\!+\!4}{2}}+\sqrt{% \frac{6\!-\!4}{2}}=1\!+\!\sqrt{5},$

and because  $4^{2}\!-\!7=9=3^{2}$,  we get

 $\sqrt{4\!-\!\sqrt{7}}=\sqrt{\frac{4\!+\!3}{2}}-\sqrt{\frac{4\!-\!3}{2}}=\frac{% \sqrt{7}\!-\!1}{\sqrt{2}}=\frac{\sqrt{14}\!-\!\sqrt{2}}{2}.$

## References

• 1 K. Väisälä: Algebran oppi- ja esimerkkikirja I.   – Werner Söderström osakeyhtiö, Porvoo & Helsinki (1952).
Title square root of square root binomial SquareRootOfSquareRootBinomial 2013-03-22 15:21:21 2013-03-22 15:21:21 pahio (2872) pahio (2872) 10 pahio (2872) Topic msc 11A25 TakingSquareRootAlgebraically SquareRootsOfRationals square root binomial square root polynomial