The Quadratic Formula can be used to solve any quadratic (or...

The Quadratic Formula can be used to solve any quadratic (or second-degree) equation. You might have wondered whether similar formulas exist for cubic (third-degree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic $x^{3}+p x+q=0,$ Cardano found the following formula for one solution: $$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$ A formula for quartic equations was discovered by the Italian mathematician Ferrari in $1540 .$ In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for fifth-degree equations. Finally, Galois gave a criterion for determining which equations can be solved by a formula involving radicals. Use the formula given above to find a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier? $$\begin{array}{l}\text { a. } x^{3}-3 x+2=0 \\\text { b. } x^{3}-27 x-54=0 \\\text { c. } x^{3}+3 x+4=0\end{array}$$

The Quadratic Formula can be used to solve any quadratic (or second-degree) equation. You might have wondered whether similar formulas exist for cubic (third-degree), quartic (fourth-degree), and higher-degree equations. For the depressed cubic $x^{3}+p x+q=0,$ Cardano found the following formula for one solution:
$$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$
A formula for quartic equations was discovered by the Italian mathematician Ferrari in $1540 .$ In 1824 the Norwegian mathematician Niels Henrik Abel proved that it is impossible to write a quintic formula, that is, a formula for fifth-degree equations. Finally, Galois gave a criterion for determining which equations can be solved by a formula involving radicals.
Use the formula given above to find a solution for the following equations. Then solve the equations using the methods you learned in this section. Which method is easier?
$$\begin{array}{l}\text { a. } x^{3}-3 x+2=0 \\\text { b. } x^{3}-27 x-54=0 \\\text { c. } x^{3}+3 x+4=0\end{array}$$

Key Concepts

Depressed Cubic Equations

A depressed cubic is a cubic equation that has been simplified to eliminate the quadratic (x²) term, taking the form x³ + p*x + q = 0. This transformation, commonly achieved via a substitution from a general cubic, simplifies analysis and is essential for methods like Cardano’s solution.

Cardano's Formula

Cardano's formula is a historic method that provides an explicit solution for depressed cubic equations by expressing one of the roots in terms of radicals. It involves taking cube roots of expressions constructed from the coefficients p and q, marking a significant early breakthrough in solving higher-degree polynomials.

Alternative Methods for Solving Cubic Equations

In certain cases, particularly when cubic equations have simple or easily recognizable roots, alternative methods like factoring, synthetic division, or graphical analysis can be more efficient and user-friendly compared to the more involved Cardano’s formula. These methods rely on pattern recognition and arithmetic simplicity rather than the formal radical approach.

Transcript

00:01 So in this question it is given to me that one of the root of a depressed cubic can be found out using this formula.

00:10 So i have a cubic given to me by x cube minus 3x plus 2 equal to 0, where if i conclude from this general cubic that is given, x cube plus px plus q equal to 0.

00:26 So comparing with this i will say p is equal to minus 3 and q is equal to 2.

00:35 So substituting the values i will get x is equal to cube root of minus q by 2, that is minus 1 plus under root of q square by 4, that is 4 by 4, that is 1 only, plus p cube by 27, that is p cube by 27 so that will be minus 1 right plus cube root of again this is minus 1 minus under root of 1 plus minus 1 so this value is clearly equal to cube root of minus 1 and this part is 0 anyways plus cube root of minus 1 so this value gives me minus 1 minus 1 minus minus 2 that is one of the root so after getting this one solution i can substitute and i can find out the other roots otherwise i can directly use synthetic division method and i can find out otherwise straight away if using normal method i had to find out i would have directly thought of some x values maybe 1 i can see x equal to 1 is one of the solution so obviously i will have the solution like x minus 1 into this one solution that is x plus 2 and another solution will be x minus 1 again.

02:15 So even on factorizing i will get this kind of roots.

02:18 So the cubic is x plus 2 into x minus 1 whole square.

02:24 So x equal to 1 is a root repeated 2 times.

02:27 For the next cubic, x cube minus 27 x minus 54.

02:42 So i can observe here that p value is equal to minus 27 and q value is equal to 54.

02:56 So remembering the formula once properly, when i substitute the values, i'll be getting x is equal to cube root of under root minus q by 2, that is minus 27 plus minus under root, that is square root of q square by 4.

03:17 So that is 54 square by 4.

03:21 So that i can write it as 27 square, all right.

03:28 Plus this one will be p cube by 27.

03:32 So that is minus 27 into minus 27 into minus 27 divided by 27.

03:40 So that will be minus 1 once again.

03:42 So effectively i will see this cube root of minus 27 plus in this.

03:55 Square root the same expression will be coming this time this will be a negative sign right this is a negative sign and a square root of 27 square and this will be minus 27 square so effectively it is zero only right so when i proceed further this value will be cube root of minus 27 plus this value is 0 plus again cube root of minus 27 and the value is zero so that that gives me minus 3 minus 3 that is equal to minus 6.

04:26 So minus 6 is one of the root...

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