Solving $x^3+24x^2+5x-13 = 0$

74 Views Asked by At

I have been doing some mathematics and I have come across this question that I need to somehow solve to move onto the next questions. I'm currently undertaking an abstract algebra unit with this questions as part of an assignment:

$x^3+24x^2+5x-13 = 0$

Can anyone help and solve with a step by step method so that I have a stronger understanding rather than just receiving the final answer?

$x_1 = 0.63204...$ $x_2 = -23.76661...$ $x_3 = -0.86543...$

Thanks! :)

1

There are 1 best solutions below

3
On BEST ANSWER

Why not to follow the steps described here to have exact and nice-looking formulae ?

For your equation $$p=-187 \qquad q=971 \qquad \implies \qquad\Delta=700105$$ so three real roots.

Now apply the trigonometric method to get $$x_k=-8+2 \sqrt{\frac{187}{3}} \cos \left(\frac{2 k\pi }{3}+\frac{1}{3} \cos ^{-1}\left(-\frac{2913 }{374}\sqrt{\frac{3}{187}}\right)\right)\qquad (k=0,1,2)$$ which give your numbers.