Algebra-23-6-6-4-2-p-q-2-then-q-2-p-2-p-2-q-

Question Number 135910 by liberty last updated on 17/Mar/21

A l g e b r a

\sqrt{23 - 6 \sqrt{6 - 4 \sqrt{2}}} = p + q \sqrt{2}

t h e n \frac{q^{2}}{p \sqrt{2}} + p^{2} q = ?

Answered by som(math1967) last updated on 17/Mar/21

\sqrt{23 - 6 \sqrt{4 + 2 - 2.2 \sqrt{2}}}

\sqrt{23 - 6 \sqrt{{(2 - \sqrt{2})}^{2}}}

\sqrt{23 - 6 \times 2 + 6 \sqrt{2}}

\sqrt{11 + 6 \sqrt{2}}

\sqrt{9 + 2 + 6 \sqrt{2}}

3 + \sqrt{2} = p + q \sqrt{2}

∴ p = 3, q = 1

\frac{q^{2}}{p \sqrt{2}} + p^{2} q = \frac{1}{3 \sqrt{2}} + 9