Question

Using the discriminant, give the nature of the roots of equation. Also solve the equation.
b) Find the cubic equation whose roots are the cubes of the roots of equation.
c) Obtain the resolvent cubics, by Descartes’ method and by Ferrari’s method, of the equation equation. Are the cubics the same? Further, use either method to obtain the roots of this equation.

09 Jan 2026
Answer :
Word Count : 273
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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