Cardano is the way to go

Algebra Level 3

Consider the following equation:

x 3 3 x 2 2 x + 4 6 2 = 0. x^3-3x^2-2x+4-6\sqrt{2}=0.

What is the real root of this equation? Round your answer to three decimal places.


The answer is 3.828.

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1 solution

Andy Hayes
Jan 23, 2017

Relevant wiki: Cardano's Method

Consider Cardano's method.

a = 1 , b = 3 , c = 2 , d = 4 6 2 . a=1, \quad b=-3, \quad c=-2, \quad d=4-6\sqrt{2}.

Compute Q Q and R : R:

Q = 3 a c b 2 9 a 2 = 5 3 R = 9 a b c 27 a 2 d 2 b 3 54 a 3 = 3 2 . \begin{aligned} Q &= \frac {3 a c - b^2} {9 a^2} \\ \\ &=-\frac{5}{3} \\ \\ \\ R &= \frac {9 a b c - 27 a^2 d - 2 b^3} {54 a^3} \\ \\ &= 3\sqrt{2}. \end{aligned}

Compute S S and T : T:

S = R + Q 3 + R 2 3 = 3 2 + 19 3 9 3 T = R Q 3 + R 2 3 = 3 2 19 3 9 3 \begin{aligned} S &= \sqrt [3] {R + \sqrt{Q^3 + R^2}} \\ \\ &= \sqrt[3]{3\sqrt{2}+\frac{19\sqrt{3}}{9}} \\ \\ \\ T &= \sqrt [3] {R - \sqrt{Q^3 + R^2}} \\ \\ &= \sqrt[3]{3\sqrt{2}-\frac{19\sqrt{3}}{9}} \end{aligned}

The real root will be:

x 1 = S + T b 3 a = 3 2 + 19 3 9 3 + 3 2 19 3 9 3 + 1 3.828 \begin{aligned} x_1 &= S + T - \frac b {3 a} \\ \\ &= \sqrt[3]{3\sqrt{2}+\frac{19\sqrt{3}}{9}} + \sqrt[3]{3\sqrt{2}-\frac{19\sqrt{3}}{9}} +1 \\ \\ &\approx \boxed{3.828} \end{aligned}

Incidentally, this value is equal to 2 2 + 1. 2\sqrt{2}+1. This shows a weakness of Cardano's method: the result tends to contain a sum of cubic roots which isn't expressed in the simplest possible way.

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