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Question Number 115597 by aurpeyz last updated on 26/Sep/20

Answered by 1549442205PVT last updated on 27/Sep/20

$$4\mathrm{a})\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$\frac{1}{\sqrt{\mathrm{x}+1}-\sqrt{\mathrm{x}}}-\frac{2}{\sqrt{\mathrm{x}-2}+\sqrt{\mathrm{x}}}=\sqrt{\mathrm{x}-1}$$$$\iff \frac{\sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}}}{1}-\frac{2(\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}-2})}{2}=\sqrt{\mathrm{x}-1}$$$$\iff \sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}}-\sqrt{\mathrm{x}}+\sqrt{\mathrm{x}-2}=\sqrt{\mathrm{x}-1}$$$$\iff \sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}-2}=\sqrt{\mathrm{x}-1}\left(1\right)$$$$\mathrm{We}\mathrm{need}\mathrm{the}\mathrm{condition}:\mathrm{x}⩾2.\mathrm{Then}$$$$\left(1\right)\iff \mathrm{x}+1+\mathrm{x}-2+2\sqrt{(\mathrm{x}+1)(\mathrm{x}-2)}=\mathrm{x}-1$$$$\iff -\mathrm{x}=2\sqrt{{\mathrm{x}}^2-\mathrm{x}-2}$$$$\mathrm{This}\mathrm{equation}\mathrm{has}\mathrm{no}\mathrm{roots}\mathrm{since}$$$$\mathrm{for}\mathrm{\forall }\mathrm{x}⩾2\Rightarrow \mathrm{LHS}<0,\mathrm{RHS}⩾0$$$$\mathrm{b})\mathrm{Prove}\mathrm{that}\mathrm{\forall }\mathrm{a},\mathrm{b}>0$$$$2(\mathrm{a}+\mathrm{b})(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2})(\frac{{\mathrm{a}}^4}{{\mathrm{b}}^2}+\frac{{\mathrm{b}}^4}{{\mathrm{a}}^2})⩾16\sqrt{\mathrm{ab}}(\ast )$$$$\mathrm{We}\mathrm{have}\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}={(\frac{1}{\mathrm{a}}-\frac{1}{\mathrm{b}})}^2+\frac{2}{\mathrm{ab}}⩾\frac{2}{\mathrm{ab}}\left(1\right)$$$$\frac{{\mathrm{a}}^4}{{\mathrm{b}}^2}+\frac{{\mathrm{b}}^4}{{\mathrm{a}}^2}={(\frac{{\mathrm{a}}^2}{\mathrm{b}}-\frac{{\mathrm{b}}^2}{\mathrm{a}})}^2+2\mathrm{ab}\left(2\right)$$$$\mathrm{a}+\mathrm{b}={\left(\sqrt{\mathrm{a}-}\sqrt{\mathrm{b}}\right)}^2+2\sqrt{\mathrm{ab}}⩾2\sqrt{\mathrm{ab}}\left(3\right)$$$$\mathrm{Multiplying}\mathrm{three}\mathrm{above}\mathrm{inequalities}$$$$\mathrm{we}\mathrm{get}2(\mathrm{a}+\mathrm{b})(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2})(\frac{{\mathrm{a}}^4}{{\mathrm{b}}^2}+\frac{{\mathrm{b}}^4}{{\mathrm{a}}^2})$$$$⩾2.\frac{2}{\mathrm{ab}}.2\mathrm{ab}.2\sqrt{\mathrm{ab}}=16\sqrt{\mathrm{ab}}.\mathrm{Hence},\mathrm{the}$$$$\mathrm{inequality}(\ast )\mathrm{is}\mathrm{proved}$$

Answered by TANMAY PANACEA last updated on 27/Sep/20

$$x^4+4x^3+11x^2+14x-30$$$$81-4\times 27+99-42-30$$$$180-108-72=0\to so(x+3)isfactor$$$$x^4+4x^3+11x^2+14x-30$$$$=x^4+3x^3+x^3+3x^2+8x^2+24x-10x-30$$$$=x^3(x+3)+x^2(x+3)+8x(x+3)-10(x+3)$$$$=(x+3)(x^3+x^2+8x-10)$$$$=(x+3)(x^3+x^2+8x-10)$$$$soonerootisx=-3$$$$★x^3+x^2+8x-10$$$$=x^3-x^2+2x^2-2x+10x-10$$$$=x^2(x-1)+2x(x-1)+10(x-1)$$$$=(x-1)(x^2+2x+10)$$$$anotherrootx=1$$$$x^2+2x+10=0$$$$x=\frac{-2\pm \sqrt{4-4\times 1\times 10}}{2\times 1}=\frac{-2\pm 6i}{2}=-1\pm 3i$$$$sorootsare(-3,1,-1\pm 3i)$$$$$$

Answered by aurpeyz last updated on 26/Sep/20

$$4a$$

Answered by TANMAY PANACEA last updated on 27/Sep/20

$$5x^5+31x^4+36x^3+36x^2+31x+5=0$$$$5x^5+5x^4+26x^4+26x^3+10x^3+10x^2+26x^2+26x+5x+5=0$$$$5x^4(x+1)+26x^3(x+1)+10x^2(x+1)+26x(x+1)+5(x+1)=0$$$$(x+1)(5x^4+26x^3+10x^2+26x+5)=0$$$$x=-1isoneroot$$$$★★now$$$$5x^4+26x^3+10x^2+26x+5=0\to devideby{x}^2$$$$5x^2+\frac{5}{x^2}+26x+\frac{26}{x}+10=0$$$$5\{{(x+\frac{1}{x})}^2-2\}+26(x+\frac{1}{x})+10=0$$$$5{(x+\frac{1}{x})}^2+26(x+\frac{1}{x})=0$$$$(x+\frac{1}{x})(5x+\frac{5}{x}+26)=0$$$$(x+\frac{1}{x})=0\to x^2+1=0x=\pm i$$$$(5x+\frac{5}{x}+26)$$$$5x^2+5+26x=0$$$$5x^2+25x+x+5=0$$$$5x(x+5)+1(x+5)=0$$$$(x+5)(5x+1)=0\to x=-5,\frac{-1}{5}$$$$sox=-1,-5,\frac{-1}{5},\pm i$$$$$$

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