How can I determine the "nature" of the roots of x^3-9*x^2+26*x-24 =0

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justaguide | College Teacher | (Level 2) Distinguished Educator

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Any cubic equation `ax^3 + bx^2 + cx + d = 0` has three roots; of these either all are real or at least one of them is real. Also, it is possible to determine whether the roots are common from the coefficients of the equation. This can be done by evaluating the expression `D = 18abcd - 4b^3d + b^2c^2 - 4ac^3 - 27a^2d^2` . If D < 0, the cubic equation has one real root and two complex conjugate roots. If D = 0, all the roots are real and there is a common root and if D > 0, there are 3 real and distinct roots.

Taking the coefficients of `x^3-9*x^2+26*x-24 =0` , `D = 18*1*-9*26*-24 + (-9)^2*26^2 - 4*1*26^3 - 27*1^2*(-24)^2 = 69988.`

As 69988 > 0, the given cubic equation has 3 distinct real roots.

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