Respuesta :
We must find the roots of the following cubic polynomial:
[tex]x^3-3x^2-7=0._{}[/tex]---------------
Theory
The roots of the cubic equation:
[tex]x^3+a_1x^2+a_2x^2+a_3=0,[/tex]are given by the following formulas:
[tex]\begin{gathered} x_1=S+T-\frac{1}{3}a_1, \\ x_2=-\frac{1}{2}(S+T)-\frac{1}{3}a_1+\frac{1}{2}i\cdot\sqrt[]{3}\cdot(S-T), \\ x_3=-\frac{1}{2}(S+T)-\frac{1}{3}a_1-\frac{1}{2}i\cdot\sqrt[]{3}\cdot(S-T)\text{.} \end{gathered}[/tex]Where Q, S and T are given by:
[tex]\begin{gathered} Q=\frac{3a_2-a^2_1}{9}, \\ R=\frac{9a_1a_2-27a_3-2a^3_1}{54}, \\ S=\sqrt[3]{R+\sqrt[]{Q^3+R^2}}, \\ T=\sqrt[3]{R-\sqrt[]{Q^3+R^2}}\text{.} \end{gathered}[/tex]The discriminant of the polynomial is:
[tex]D=Q^3+R^2.[/tex]We have the following possibilities:
0. one root is real and two complex conjugates if D > 0,
,1. all roots are real and at last two are equal if D = 0,
,2. all roots are real and unequal if D < 0.
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We rewrite the polynomial of the problem in the following way:
[tex]x^3+(-3)\cdot x^2+0\cdot x+(-7)=0._{}[/tex]Comparing the last equation with the general case, we identify the coeffients:
[tex]\begin{gathered} a_1=-3, \\ a_2=0, \\ a_3=-7. \end{gathered}[/tex]Replacing the values of the coefficients in the formulas of Q and R, we get:
[tex]\begin{gathered} Q=\frac{3\cdot0-(-3)^2}{9}=-1, \\ R=\frac{9\cdot(-3)\cdot0-27\cdot(-7)-2\cdot(-3)^2}{54}=\frac{9}{2}. \end{gathered}[/tex]Replacing the values of Q and R in the formula of S and T, we get:
[tex]\begin{gathered} S=\sqrt[3]{(\frac{9}{2})+\sqrt[]{(-1)^3+(\frac{9}{2})^2}}=\sqrt[]{\frac{9}{2}+\frac{\sqrt[]{77}}{2}}, \\ T=\sqrt[3]{(\frac{9}{2})-\sqrt[]{(-1)^3+(\frac{9}{2})^2}}=\sqrt[]{\frac{9}{2}-\frac{\sqrt[]{77}}{2}}\text{.} \end{gathered}[/tex]The discriminant of the polynomial is:
[tex]D=Q^3+R^2=(-1)^3+(\frac{\sqrt[]{77}}{2})^2=\frac{73}{4}>0.^{}[/tex]We have D > 0 so one root is real and two complex conjugates.
We compute only the real root, which is given by:
[tex]\begin{gathered} x_1=S+T-\frac{1}{3}a_1, \\ x_1=\sqrt[]{\frac{9}{2}+\frac{\sqrt[]{77}}{2}}+\sqrt[]{\frac{9}{2}-\frac{\sqrt[]{77}}{2}}-\frac{1}{3}(-3), \\ x_1=\sqrt[]{\frac{9}{2}+\frac{\sqrt[]{77}}{2}}+\sqrt[]{\frac{9}{2}-\frac{\sqrt[]{77}}{2}}+1, \\ x_1\cong3.55415. \end{gathered}[/tex]Answer
The cubic equation has one real root and two complex roots. The real root is:
[tex]x_1\cong3.55415.[/tex]