Question-59600

Question Number 59600 by Gulay last updated on 12/May/19

Answered by MJS last updated on 12/May/19

(a+b(√3))^2 =14+3(√3) a^2 +3b^2 +2ab(√3)=14+3(√3) ⇒ a^2 +3b^2 =14 2ab=3 ⇒ b=(3/(2a)) a^2 +((27)/(4a^2 ))=14 a^4 −14a^2 +((27)/4)=0 ⇒ a^2 =(1/2) ∨ a^2 =((27)/2) ⇒ a=±((√2)/2) ∨ a=±((3(√6))/2) ⇒ b=±((3(√2))/2) ∨ b=±((√6)/6) ⇒ (√(14+3(√3)))=((√2)/2)(1+3(√3)) ⇒ (√(14−3(√3)))=((√2)/2)(−1+3(√3)) ⇒ (√(14+3(√3)))−(√(14−3(√3)))=(√2)

$$\left({a}+{b}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{ab}\sqrt{\mathrm{3}}=\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$${a}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{2}} =\mathrm{14} \\ $$$$\mathrm{2}{ab}=\mathrm{3}\:\Rightarrow\:{b}=\frac{\mathrm{3}}{\mathrm{2}{a}} \\ $$$${a}^{\mathrm{2}} +\frac{\mathrm{27}}{\mathrm{4}{a}^{\mathrm{2}} }=\mathrm{14} \\ $$$${a}^{\mathrm{4}} −\mathrm{14}{a}^{\mathrm{2}} +\frac{\mathrm{27}}{\mathrm{4}}=\mathrm{0}\:\Rightarrow\:{a}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\:\vee\:{a}^{\mathrm{2}} =\frac{\mathrm{27}}{\mathrm{2}} \\ $$$$\Rightarrow\:{a}=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\vee\:{a}=\pm\frac{\mathrm{3}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\Rightarrow\:{b}=\pm\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{2}}\:\vee\:{b}=\pm\frac{\sqrt{\mathrm{6}}}{\mathrm{6}} \\ $$$$\Rightarrow\:\sqrt{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\sqrt{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(−\mathrm{1}+\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\:\sqrt{\mathrm{14}+\mathrm{3}\sqrt{\mathrm{3}}}−\sqrt{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{3}}}=\sqrt{\mathrm{2}} \\ $$

Commented by Kunal12588 last updated on 12/May/19

sir why (√(14−3(√3)))≠((√2)/2)(1−3(√3)) is it bcoz ((√2)/2)(1−3(√3))<0?

$${sir}\:{why}\:\sqrt{\mathrm{14}−\mathrm{3}\sqrt{\mathrm{3}}}\neq\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{3}\sqrt{\mathrm{3}}\right) \\ $$$${is}\:{it}\:{bcoz}\:\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{3}\sqrt{\mathrm{3}}\right)<\mathrm{0}? \\ $$

Commented by MJS last updated on 12/May/19

(√x)>0 for x>0 (x)^(1/n) is always unique for x∈C it′s not an equation! z=re^(iθ) (z)^(1/n) =(r)^(1/n) e^(i(θ/n))

$$\sqrt{{x}}>\mathrm{0}\:\mathrm{for}\:{x}>\mathrm{0} \\ $$$$\sqrt[{{n}}]{{x}}\:\mathrm{is}\:\mathrm{always}\:\mathrm{unique}\:\mathrm{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{an}\:\mathrm{equation}! \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$$\sqrt[{{n}}]{{z}}=\sqrt[{{n}}]{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{{n}}} \\ $$

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