Finding the real zeroes of a cubic?
- Thread starter parabolas
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There is a way to track them down explicitly, but you probably don't want to know it. Normally, it is found a simpler task just to track them down numerically. Can you formulate a plan for that? What do you know about "Rule of Signs", "Rational Roots", and "Synthetic Division"? There might be a place to start in there, somewhere.
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What type of math are you doing - high school? college?
The high school method of employing rational root theorem could be a very long process here - since 720 has many factors.
Quickest method would be to use your graphical calculator - graph the function. Estimate the roots - then refine those as needed.
This given function does not have rational roots.
The high school method of employing rational root theorem could be a very long process here - since 720 has many factors.
Quickest method would be to use your graphical calculator - graph the function. Estimate the roots - then refine those as needed.
This given function does not have rational roots.
You could "complete the cube" (not really):
\(\displaystyle \L\ x^3 - 30x^2 + 275x - 720 = (x - 10)^3 - 25x + 280\)
Then put y = x - 10:
\(\displaystyle \L\ y^3 - 25y + 30 = 0\)
Applying Cardano's method, you obtain the exact solution:
\(\displaystyle \L\ y = [(\frac{5}{9}\ \sqrt{1146}\)i - 15]^{\frac{1}{3}} + \frac{25}{3}\ [(\frac{5}{9}\ \sqrt{1146}\)i - 15]^{\frac{-1}{3}}\)
Beware, you must be VERY well versed in complex numbers to be able to resolve the above expression (which contains non-real numbers) into the 3 real solutions of this cubic (as there are 3 cubic roots for any complex number). I wouldn't bother.
Using a calculator gives three approximate cubic roots of i - 15\) to be 2.1161 + 1.96353i, 0.642421 - 2.81436i, -2.75852 + 0.850828i. The reciprocal of these cube roots are 0.253932 - 0.235624i, 0.077091 + 0.337723i and -0.331022 - 0.102099i respectively. It turns out that the imaginary parts "vanish" (a phenomenon discovered by Bombelli). So, the roots for y are ~ 1.2848, 4.2322 and -5.51704. It's instances like this where imaginary numbers come in handy.
\(\displaystyle \L\ x^3 - 30x^2 + 275x - 720 = (x - 10)^3 - 25x + 280\)
Then put y = x - 10:
\(\displaystyle \L\ y^3 - 25y + 30 = 0\)
Applying Cardano's method, you obtain the exact solution:
\(\displaystyle \L\ y = [(\frac{5}{9}\ \sqrt{1146}\)i - 15]^{\frac{1}{3}} + \frac{25}{3}\ [(\frac{5}{9}\ \sqrt{1146}\)i - 15]^{\frac{-1}{3}}\)
Beware, you must be VERY well versed in complex numbers to be able to resolve the above expression (which contains non-real numbers) into the 3 real solutions of this cubic (as there are 3 cubic roots for any complex number). I wouldn't bother.
Using a calculator gives three approximate cubic roots of i - 15\) to be 2.1161 + 1.96353i, 0.642421 - 2.81436i, -2.75852 + 0.850828i. The reciprocal of these cube roots are 0.253932 - 0.235624i, 0.077091 + 0.337723i and -0.331022 - 0.102099i respectively. It turns out that the imaginary parts "vanish" (a phenomenon discovered by Bombelli). So, the roots for y are ~ 1.2848, 4.2322 and -5.51704. It's instances like this where imaginary numbers come in handy.