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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
Algebra     (√(23−6(√(6−4(√2))))) = p+q(√2)  then (q^2 /(p(√2))) + p^2 q =?
$${Algebra}\: \\ $$$$\:\:\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:{p}+{q}\sqrt{\mathrm{2}} \\ $$$${then}\:\frac{{q}^{\mathrm{2}} }{{p}\sqrt{\mathrm{2}}}\:+\:{p}^{\mathrm{2}} {q}\:=? \\ $$
Answered by som(math1967) last updated on 17/Mar/21
(√(23−6(√(4+2−2.2(√2)))))  (√(23−6(√((2−(√2))^2 ))))  (√(23−6×2+6(√2)))  (√(11+6(√2)))  (√(9+2+6(√2)))  3+(√2)=p +q(√2)  ∴p=3 ,  q=1   (q^2 /(p(√2))) +p^2 q=(1/(3(√2))) +9
$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{4}+\mathrm{2}−\mathrm{2}.\mathrm{2}\sqrt{\mathrm{2}}}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}×\mathrm{2}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{9}+\mathrm{2}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}={p}\:+{q}\sqrt{\mathrm{2}} \\ $$$$\therefore{p}=\mathrm{3}\:,\:\:{q}=\mathrm{1} \\ $$$$\:\frac{{q}^{\mathrm{2}} }{{p}\sqrt{\mathrm{2}}}\:+{p}^{\mathrm{2}} {q}=\frac{\mathrm{1}}{\mathrm{3}\sqrt{\mathrm{2}}}\:+\mathrm{9} \\ $$

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