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

$$usepertfectlysquarenumberstofind$$$$\sqrt{11}$$$$nocalculatorsallowed$$

Commented by Linton last updated on 09/Oct/22

$$a^{\frac{1}{2}}=\frac{a+b}{2\sqrt{b}}wherebisaperfectsquare$$

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

$$3^2+{\left(\sqrt{2}\right)}^2={(3+\sqrt{2})}^2-6\sqrt{2}$$$$={(3-2\sqrt{2})}^2-18$$$$=[{(3-2\sqrt{2}-9]}^2-9^2$$$$=2^2{(3+\sqrt{2})}^2-9^2$$

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