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Question Number 100146 by ajfour last updated on 25/Jun/20

Commented by bemath last updated on 25/Jun/20

$$\mathrm{t}=\frac{{\mathrm{c}}^2-2\mathrm{b}-1}{2(\mathrm{b}+1)}$$

Commented by ajfour last updated on 25/Jun/20

$$clearly\sqrt{b+t}=\frac{c}{t}$$$$or(b+t)t^2=c^2$$$$letmecheck$$$$(b+\frac{c^2-2b-1}{2(b+1)}){\left[\frac{c^2-2b-1}{2(b+1)}\right]}^2$$$$\ne c^2$$$$pleasecheckyouranswersir,$$$$orpostyourworkings..$$

Commented by ajfour last updated on 25/Jun/20

$$Findtintermsofb,c.$$

Answered by mr W last updated on 25/Jun/20

$$\frac{y}{1}=\frac{b+t}{y}=\frac{c}{t}$$$$\Rightarrow {\left(\frac{c}{t}\right)}^2=b+t$$$$\Rightarrow t^3+{bt}^2-c^2=0$$$$...$$

Commented by ajfour last updated on 25/Jun/20

$$thanlssir,ihopesomeday$$$$somegeometryshallpavethe$$$$way..!$$

Commented by bemath last updated on 25/Jun/20

$$\mathrm{where}\mathrm{becomes}\mathrm{y}?$$

Commented by 1549442205 last updated on 25/Jun/20

$$\mathrm{Yes},\mathrm{sir}\mathrm{Mr}.\mathrm{W}.\mathrm{Please},\mathrm{let}\mathrm{me}\mathrm{be}\mathrm{continued}.$$$$\mathrm{Putting}\mathrm{x}=\frac{1}{\mathrm{t}}\mathrm{we}\mathrm{get}{\mathrm{c}}^2{\mathrm{x}}^3-\mathrm{bx}-1=0$$$$\iff {\left(\mathrm{cx}\right)}^3-\mathrm{bcx}-\mathrm{c}=\mathrm{o}.\mathrm{Put}\mathrm{cx}=\mathrm{u}\mathrm{we}\mathrm{get}$$$${\mathrm{u}}^3-\mathrm{bu}-\mathrm{c}=0\left(1\right).\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{roots}\mathrm{of}$$$$\left(1\right)\mathrm{in}\mathrm{the}\mathrm{form}:\mathrm{u}=-(\mathrm{m}+\mathrm{n}).\mathrm{Then}$$$$\mathrm{u}+\mathrm{m}+\mathrm{n}=0\Rightarrow {\mathrm{u}}^3+{\mathrm{m}}^3+{\mathrm{n}}^3-3\mathrm{umn}=0\left(2\right)$$$$\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get}\{\begin{array}{l}{\mathrm{m}}^3+{\mathrm{n}}^3=-\mathrm{c}\\ \mathrm{mn}=\frac{\mathrm{b}}{3}\end{array}$$$$\Rightarrow {({\mathrm{m}}^3-{\mathrm{n}}^3)}^2={({\mathrm{m}}^3+{\mathrm{n}}^3)}^2-{4\mathrm{m}}^3{\mathrm{n}}^3={\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}$$$$\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{have}\mathrm{the}\mathrm{condition}\mathrm{that}$$$${\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}⩾0\iff \mathrm{c}⩾\frac{2\mathrm{b}}{9}\sqrt{3\mathrm{b}}.\mathrm{Then}\mathrm{we}\mathrm{get}$$$$\{\begin{array}{l}{\mathrm{m}}^3+{\mathrm{n}}^3=-\mathrm{c}\\ {\mathrm{m}}^3-{\mathrm{n}}^3=\sqrt{{\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}}\end{array}\Rightarrow \{\begin{array}{l}\mathrm{m}={}^3\sqrt{\frac{-\mathrm{c}+\sqrt{{\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}}}{2}}\\ \mathrm{n}={}^3\sqrt{\frac{-\mathrm{c}-\sqrt{{\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}}}{2}}\end{array}$$$$\Rightarrow \mathrm{u}=-(\mathrm{m}+\mathrm{n})={}^3\sqrt{\frac{\mathrm{c}-\sqrt{{\mathrm{c}}^2-\frac{{4\mathrm{b}}^3}{27}}}{2}}+{}^3\sqrt{\frac{\mathrm{c}+\sqrt{{\mathrm{c}}^2-\frac{{4\mathrm{b}}^2}{27}}}{2}}$$$$\mathbf{t}=\frac{1}{\mathbf{x}}=\frac{\mathbf{c}}{\mathbf{u}}=\frac{\mathbf{c}}{3\sqrt{\frac{\mathbf{c}-\sqrt{{\mathbf{c}}^2-\frac{4{\mathbf{b}}^3}{27}}}{2}}+{}^3\sqrt{\frac{\mathbf{c}+\sqrt{{\mathbf{c}}^2-\frac{4{\mathbf{b}}^3}{27}}}{2}}}$$$$$$

Commented by mr W last updated on 25/Jun/20

$$thankyou,verynicesir!$$$$ihavehadappliedcardanodirectly.$$

Commented by mr W last updated on 25/Jun/20

Commented by ajfour last updated on 25/Jun/20

$$thanks....205Sir,butcannot$$$$wehaveananswerforany$$$$realbamdc?$$

Commented by 1549442205 last updated on 26/Jun/20

$$\mathrm{There}\mathrm{is}\mathrm{always}\mathrm{exists}\mathrm{some}\mathrm{root}\mathrm{for}$$$$\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{so}\mathrm{that}{\mathrm{c}}^2⩾\frac{{4\mathrm{b}}^3}{27}.$$$$\mathbf{Example},\mathbf{b}=3,\mathbf{c}=4\Rightarrow \mathbf{t}\approx 1.821640$$

Commented by 1549442205 last updated on 30/Jun/20

$$\mathrm{But}\mathrm{I}\mathrm{want}\left(\mathrm{please}\mathrm{get}\mathrm{his}\mathrm{permission}\right)$$$$\mathbf{to}\mathbf{do}\mathrm{the}\mathrm{his}\mathrm{work}$$$$\mathrm{again}\mathrm{in}...\mathrm{the}\mathrm{moment}!.!.!$$

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