Question
Using the discriminant, give the nature of the roots of . Also solve the equation.
b) Find the cubic equation whose roots are the cubes of the roots of .
c) Obtain the resolvent cubics, by Descartes’ method and by Ferrari’s method, of the equation . Are the cubics the same? Further, use either method to obtain the roots of this equation.
Answer :
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Skip in 10
Consider first the equation (7x^{3}+x^{2}-35x-5=0). Dividing by 7 gives (x^{3}+\frac{1}{7}x^{2}-5x-\frac{5}{7}=0). For a cubic (x^{3}+px^{2}+qx+r=0), the discriminant is (\Delta=18pqr-4p^{3}r+p^{2}q^{2}-4q^{3}-27r^{2}). Here (p=\frac17,; q=-5,; r=-\frac57). Substituting, [ \Delta =18!\left(\frac17\right)(-5)!\left(-\frac57\right) -4!\left(\frac17\right)^{3}!\left(-\frac57\right) +\left(\frac17\right)^{2}(25) -4(-125) -27!\left(\frac{25}{49}\right). ] Simplifying term by term gives [ \Delta=\frac{450}{49}+\frac{20}{343}+\frac{25}{49}+500-\frac{675}{49}. ] Combining the fractional terms, [ \frac{450+25-675}{49}+\frac{20}{343} =-\frac{200}{49}+\frac{20}{343} =-\frac{1380}{343}. ] Hence [ \Delta=500-\frac{1380}{343}=\frac{170120-1380}{343} =\frac{168740}{343}>0. ] Therefore the equation has three distinct real roots. Trying simple rational roots in the original equation, (x=1) gives (7+1-35-5=-32\neq0), while (x=5) gives (875+25-175-5\neq0).
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