use-pertfectly-square-numbers-to-find-11-no-calculators-allowed-

Question Number 177768 by Linton last updated on 08/Oct/22

u s e p e r t f e c t l y s q u a r e n u m b e r s t o f i n d

\sqrt{11}

n o c a l c u l a t o r s a l l o w e d

Commented by Linton last updated on 09/Oct/22

a^{\frac{1}{2}} = \frac{a + b}{2 \sqrt{b}} w h e r e b i s a p e r f e c t s q u a r e

Answered by a.lgnaoui last updated on 09/Oct/22

3^{2} + {(\sqrt{2})}^{2} = {(3 + \sqrt{2})}^{2} - 6 \sqrt{2}

= {(3 - 2 \sqrt{2})}^{2} - 18

= [{(3 - 2 \sqrt{2} - 9]}^{2} - 9^{2}

= 2^{2} {(3 + \sqrt{2})}^{2} - 9^{2}