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Question Number 36132 by Rasheed.Sindhi last updated on 29/May/18

$$a+b=10.........\left(\mathrm{i}\right)$$$$ab+c=0..........\left(\mathrm{ii}\right)$$$$ac+d=6..........\left(\mathrm{iii}\right)$$$$ad=-1...........\left(\mathrm{iv}\right)$$$$(a,b,c,d)=?$$$$\mathcal{N}ote:Thisproblemisrelatedtosolve$$$$theequation(t^4+10t+6t-1=0)of$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Commented by Rasheed.Sindhi last updated on 29/May/18

$$\left(\mathrm{i}\right)\Rightarrow a=10-b\Rightarrow$$$$\left(\mathrm{ii}\right):(10-b)b+c=0\Rightarrow c=(b-10)b$$$$\left(\mathrm{iii}\right):(10-b)c+d=6\Rightarrow (10-b)\left\{(b-10)b\right\}=6$$$$\left(\mathrm{iv}\right):(10-b)d=-1$$$$\left(\mathrm{ii}\right):c=(b-10)b\Rightarrow$$$$\left(\mathrm{iii}\right):(10-b)(b-10)b-\frac{1}{10-b}=6$$$$-(b-10)(b-10)b+\frac{1}{b-10}=6$$$$-b{(b-10)}^3+1=6(b-10)$$$$$$$$W\mathrm{rongly}\mathrm{derived}\mathrm{cubic}\mathrm{equation}.\mathrm{Sorry}!$$$$b^3-20b^2+100b+6=0$$$$A\mathrm{unybody}\mathrm{who}\mathrm{know}\mathrm{how}\mathrm{to}\mathrm{solve}\mathrm{cubic}\mathrm{equation}..$$

Commented by prakash jain last updated on 29/May/18

$$\mathrm{There}\mathrm{is}\mathrm{a}\mathrm{procedure}\mathrm{to}\mathrm{solve}\mathrm{cubic}$$$$\mathrm{equation}\mathrm{and}\mathrm{also}\mathrm{cubic}\mathrm{formula}.$$$$$$$$\mathrm{It}\mathrm{is}\mathrm{easier}\mathrm{to}\mathrm{follow}\mathrm{step}\mathrm{then}\mathrm{remembering}$$$$\mathrm{the}\mathrm{formula}.$$

Commented by mondodotto@gmail.com last updated on 29/May/18

$$\mathrm{use}\mathrm{a}\mathrm{calculator}$$

Commented by prakash jain last updated on 29/May/18

http://www.sosmath.com/algebra/factor/fac11/fac11.html

Commented by Rasheed.Sindhi last updated on 29/May/18

$$\mathcal{T}h\mathrm{\infty }nk\mathcal{S}\mathcal{S}ir!Ialwayshavelearnt$$$$fromyou.$$

Commented by MJS last updated on 29/May/18

$$\mathrm{the}\mathrm{problem}\mathrm{with}{\mathrm{Cardano}}^{\prime }\mathrm{s}\mathrm{method}\mathrm{is},$$$$\mathrm{you}{\mathrm{won}}^{\prime }\mathrm{t}\mathrm{get}\mathrm{real}\mathrm{solutions}\mathrm{of}\mathrm{usable}\mathrm{form}$$$$(x-3)(x+4)(x-5)=0$$$$x^3-4x^2-17x+60=0$$$$x=z+\frac{4}{3}$$$$z^3-\frac{67}{3}z+\frac{880}{27}=0$$$$p=-\frac{67}{3};q=\frac{880}{27}$$$$z_1=u+v$$$$z_2=(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})v$$$$z_3=(-\frac{1}{2}-\frac{\sqrt{3}}{2}\mathrm{i})u+(-\frac{1}{2}+\frac{\sqrt{3}}{2}\mathrm{i})v$$$$u=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}=\sqrt[3]{-\frac{440}{27}+7\sqrt{3}\mathrm{i}}$$$$v=\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{p^3}{27}+\frac{q^2}{4}}}=\sqrt[3]{-\frac{440}{27}+7\sqrt{3}\mathrm{i}}$$$$u+v=\sqrt[3]{-\frac{440}{27}+7\sqrt{3}\mathrm{i}}+\sqrt[3]{-\frac{440}{27}+7\sqrt{3}\mathrm{i}}=\frac{11}{3}$$$$\mathrm{but}\mathrm{you}\mathrm{cannot}\mathrm{easily}\mathrm{show}\mathrm{this}\mathrm{is}\mathrm{true}$$$$\mathrm{and}\mathrm{in}\mathrm{a}\mathrm{case}\mathrm{like}$$$$(x-a-b\mathrm{i})(x-a+b\mathrm{i})(x-c)=0$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{impossible}$$

Commented by Rasheed.Sindhi last updated on 30/May/18

$$\mathcal{T}hank\mathcal{S}alot\mathcal{S}ir!$$

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