DISCUSS: The Cubic Formula The Quadratic Formula can be used...

DISCUSS: The Cubic Formula 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 (page 328) found the following formula for one solution: $$\quad x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$ 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 (page 313) 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 using equations. Then solve the equations using the methods you learned in this section. Which method is easier? (a) $x^{3}-3 x+2=0$ (b) $x^{3}-27 x-54=0$ (c) $x^{3}+3 x+4=0$

DISCUSS: The Cubic Formula 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 (page 328) found the following formula for one solution:
$$\quad x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$$
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 (page 313) 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 using equations. Then solve the equations using the methods you learned in this section. Which method is easier?
(a) $x^{3}-3 x+2=0$
(b) $x^{3}-27 x-54=0$
(c) $x^{3}+3 x+4=0$

Key Concepts

Cubic Formula (Cardano's Formula)

This is an explicit formula that provides the roots of any depressed cubic equation using radicals. It extends the idea of the quadratic formula to third-degree polynomials by expressing the solution in terms of cube roots and square roots, allowing one to solve certain cubic equations that are transformed into a simplified form without the quadratic term.

Depressed Cubic Transformation

This concept involves reducing a general cubic equation to one without a quadratic term by a suitable change of variables. The depressed form simplifies the structure of the cubic equation, making it more amenable to solution techniques such as Cardano's Formula, and highlights the underlying symmetry in the equation.

Radical Solutions of Equations

Radical solutions refer to expressing the roots of polynomial equations in terms of radicals—expressions involving nth roots. This concept is central in understanding which types of equations can be solved algebraically and is exemplified by formulas for quadratic, cubic, and quartic equations.

Abel–Ruffini Theorem

This theorem establishes that there is no general solution in radicals for polynomial equations of degree five or higher. It highlights the limitations of extending techniques like those used for quadratic, cubic, and quartic equations to all higher-order polynomials, marking a fundamental boundary in the theory of equations.

Galois Theory

Galois Theory provides a deep connection between field theory and group theory, offering criteria that determine whether a given polynomial equation can be solved by radicals. It explains the solvability of equations by analyzing the structure of their symmetry groups and underpins the modern understanding of why certain equations lack closed-form radical solutions.

Transcript

00:01 Okay, this time they talk with us about a formula for cubic equations, specifically of the form x -cube plus px plus cube is equal to zero.

00:18 Okay.

00:19 We have a quadratic formula for any quadratic that would be a x squared plus vxx plus c, and we have the quadratic.

00:33 Formula to solve this.

00:35 This time they introduced to us an equation to get exactly one solution from this.

00:40 It'll give you one.

00:41 And it's a little more complicated.

00:44 It's partially why you don't really see this in your classes.

00:56 There actually exist some formulas for fourth degree polynomials, but those are just so tremendous.

01:03 There's no way they're going to put this even in a textbook problem.

01:07 So this is as big as it will get and you'll study more about it if you're curious.

01:15 When you take more advanced algebra courses in the future.

01:20 Okay, although the perspective changes a little bit.

01:23 That's pretty cool.

01:26 Let's see, q2 and square root, q squared over four plus b squared over four.

01:38 Okay, so 20 over 27.

01:40 So this is the formula for solving depressed quintics, or depressed cubic, sorry, that's cubic of this form.

01:51 Notice there's no x squared term.

01:57 So actually there's a way to transform it, any cubic, to make it look quite like this.

02:07 And we can use this formula after transforming it to this to find one solution.

02:15 Notice how like the quadratic formula is way nicer.

02:18 It gives you all of them if they exist.

02:20 It also gives you the complex as well.

02:24 What do they want us to do? well, they give us x cubed minus 3x plus 2.

02:34 Okay, the wording is a little funny.

02:39 They want us to find a solution given the formula, and then they want us to solve the equation using the methods we learned in this section.

02:49 And they ask us which method is easier.

02:53 Okay, a little bit of a flawed question because which method is easier.

02:56 Well, one is just finding a single solution.

02:59 And one is going through the process of finding all the solutions.

03:03 And really, they can.

03:03 Complement each other.

03:04 So x cubed minus the x plus two.

03:11 Let's use our formula here.

03:15 Our one solution will be of the full.

03:19 We're going to take the cube root of let's see q's over here.

03:25 So negative q 2 over 2, negative 1 plus the square root of q squared over 4.

03:34 So that's just 4 over 4, which is 1 plus p cubed.

03:41 But that's negative 3.

03:42 So negative 3.

03:43 Cube is negative 27.

03:44 Let's say subtract one in that case, plus the cube root of, well, it's still a negative q over 2, so it's the same as this value.

03:55 This time it's minus the same inside term here.

03:58 You notice this inner square root is the same as this inner square root.

04:06 So it'll just be minus the square root of 1 minus.

04:11 Let's evaluate this.

04:12 Well, this is 0.

04:14 This is 0.

04:15 So negative 1 plus 0, negative 1 minus 0.

04:19 What do we end up with? well, the key root of negative 1 is just negative 1, and there's another negative 1.

04:27 Okay, we found one of our roots.

04:30 So we actually have the polynomial.

04:35 Let's divide it with synthetic division.

04:39 We'll use negative 2, and the coefficients are 1, 0, negative 3, and 2.

04:44 Okay, all the form synthetic division, multiply these two numbers, add going down, multiply this number, this number, add going down for it and then okay multiply this number this number add going down you get zero so the polynomial we would get after that is x plus two x squared minus two x plus one okay well what is this this is just x minus one squared i believe you can foil it out so squared so there are exactly two solutions to this question.

05:24 Negative two and one.

05:30 Okay, let's see which method is easier.

05:34 Well, one complements the other.

05:35 I can foresee scenarios where the numbers aren't this nice, and this is just horrible to compute, and it might have been easier to ignore it and use the synthetic division.

05:47 But i feel like this could give you irrational numbers as well.

05:50 We wouldn't have gotten otherwise.

05:54 There are some scenarios where it's probably necessary, but overall, if it's possible, to use both methods and this one isn't necessary, it's probably faster just to use synthetic.

06:04 But not all problems will be carefully designed and given to you for this book.

06:08 So let's keep in mind that this formula exists in the future if it ever comes in him.

06:15 Let's see.

06:18 Let's go for actually let me see if i can move this work.

06:24 Because the formulas are right there.

06:25 I'd like to not use them.

06:27 No, i know.

06:30 I just want to grab mine.

06:35 Okay.

06:36 I remove some of my circles now let's do part b they give us the cubic x cubed minus 27 x 24 is equal to zero okay so again we're going to open with our cubic formula over here we will end up with let's see x is equal to the cube root of negative q so that would just be 54 over 2 but that should be 27.

07:26 If they can multiply 27 by 2 you get 40 plus 14 54 okay plus the square root of q squared well q squared in this case is 54 squared it's 27 times 2 squared both of them are squared over 4.

07:53 Okay.

07:57 Plus p cubed over 27.

08:00 But b cubed is already 27.

08:04 There's a negative.

08:05 So this becomes negative.

08:08 27 cubed over 27.

08:12 So one of the 27s cancels out.

08:14 This is just 27 squared...