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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

2x^3 + x = 171

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

I agree. But just playing a hunch should this be

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

Not a mistake ! So do you mean it can't be solved ?

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

hi Abbey,

It has one real solution and two complex ones. You can use the MIF function grapher

https://www.mathsisfun.com/data/function-grapher.php

to find an approximate real solution.

Where did the question come from? What are you studying? Telling us this will help us to help you.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

It was from no where i was just thinking through some complex equations and i found this but i think we can solve it as a cubic equation in this way 2x^3 + 0x^2 + x = 171

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

I have used an online cubic equation solver giving a = 2 b = 0 c = 1 d = -171 and got 3 solutions

[-2.1838+3.84798i]

[-2.1838-3.84798i]

[4.3676] try proving it

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

Once the real solution is known the remaining two can be found by extracting the quadratic and using the formula. But if you didn't know it, you'd have to use the cubic formula which is quite complicated.

You can also use https://www.wolframalpha.com/

The maths software there will solve many equations. Type 'solve followed by the equation'

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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As an alternative, the equation is a depressed cubic -- so if we let and compare the resulting equation with the identitythen this gives us two equations in and which can be solved using the quadratic formula. Adding together and then gives you a solution to the original cubic.

More generally any cubic can be reduced to a depressed cubic by substitution -- and then the method above can be applied.

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

Wait i have heard of a cubic formula! What is it ¿¿¿¿

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 45,477

Hi Abbey78336,

**Cubic equation**

In algebra, a cubic equation in one variable is an equation of the form

in which a is nonzero.

The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a, b, c, and d of the cubic equation are real numbers, then it has at least one real root (this is true for all odd-degree polynomial functions). All of the roots of the cubic equation can be found by the following means:

* algebraically, that is, they can be expressed by a cubic formula involving the four coefficients, the four basic arithmetic operations and nth roots (radicals). (This is also true of quadratic (second-degree) and quartic (fourth-degree) equations, but not of higher-degree equations, by the Abel–Ruffini theorem.)

* trigonometrically

* numerical approximations of the roots can be found using root-finding algorithms such as Newton's method.

The coefficients do not need to be real numbers. Much of what is covered below is valid for coefficients in any field with characteristic other than 2 and 3. The solutions of the cubic equation do not necessarily belong to the same field as the coefficients. For example, some cubic equations with rational coefficients have roots that are irrational (and even non-real) complex numbers.

**Factorization**

If the coefficients of a cubic equation are rational numbers, one can obtain an equivalent equation with integer coefficients, by multiplying all coefficients by a common multiple of their denominators. Such an equation

with integer coefficients, is said to be reducible if the polynomial of the left-hand side is the product of polynomials of lower degrees. By Gauss's lemma, if the equation is reducible, one can suppose that the factors have integer coefficients.

Finding the roots of a reducible cubic equation is easier than solving the general case. In fact, if the equation is reducible, one of the factors must have degree one, and thus have the form

with q and p being coprime integers. The rational root test allows finding q and p by examining a finite number of cases (because q must be a divisor of a, and p must be a divisor of d).

Thus, one root is

and the other roots are the roots of the other factor, which can be found by polynomial long division. This other factor is(The coefficients seem not to be integers, but must be integers if p / q is a root.)

Then, the other roots are the roots of this quadratic polynomial and can be found by using the quadratic formula.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

zetafunc: many thanks for that method. It took me a while to debug my workings but eventually I got there ie. 4.3676....

I was getting depressed ( is that why it's so called?) when it kept going wrong but I'm over that now

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 45,477

Abbey78336 wrote:

Wait i have heard of a cubic formula! What is it ¿¿¿¿

Please see the post #11 above!

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

Is there a formula like a the quadratic formula where i just plug in the a,b....y,z

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 45,477

Hi Abbey78336,

**Quadratic formula**

In elementary algebra, the quadratic formula is a formula that provides the solution(s) to a quadratic equation. There are other ways of solving a quadratic equation instead of using the quadratic formula, such as factoring (direct factoring, grouping, AC method), completing the square, graphing and others.

Given a general quadratic equation of the form

with x representing an unknown, with a, b and c representing constants, and with a ≠ 0, the quadratic formula is:

where the plus–minus symbol "±" indicates that the quadratic equation has two solutions. Written separately, they become:

Each of these two solutions is also called a root (or zero) of the quadratic equation. Geometrically, these roots represent the x-values at which any parabola, explicitly given as

crosses the x-axis.As well as being a formula that yields the zeros of any parabola, the quadratic formula can also be used to identify the axis of symmetry of the parabola, and the number of real zeros the quadratic equation contains.

The expression

is known as discriminant. If then the square root of the discriminant will be a real number; otherwise it will be a complex number.If

, b, and c are real numbers then* If

then we have two distinct real roots/solutions.* If then we have one repeated real solution.

* If then we have two distinct complex solutions, which are complex conjugates of each other.

It appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Abbey78336****Member**- Registered: 2022-08-15
- Posts: 37

You are not helping me all i want is the cubic formula but you are just giving other explanations i have been studying equations of math for 6 months i know almost everything other than the cubic formula pleaseeeeeeeeee!! I need the cubic formula if it is not there lets make it maths is made by p'ple like us if you can't it means you are ./.?.¡.\.:.".. Nothing

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**Jai Ganesh****Administrator**- Registered: 2005-06-28
- Posts: 45,477

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,903

Hi Abbey,

I think Ganesh just misunderstood what you wanted. No need to be rude, he's doing his best

You now have 3 techniques for solving a cubic

As you can see it's not quite as simple as the quadratic formula.

In the most general case you'll need two substitutions to get a solution.

Zetafunc's method worked because there was no x squared term

Stefy's shows how you can get any cubic into that form.

So I'd recommend you try Zetafunc's method first on your earlier problem. He didn't put in every step (he likes to stretch us) so post again if you need more help with that.

Then you can advance to a more general cubic with Stefy's method

In theory you could feed both the substitutions into a single formula but it would be complicated. If you can write programs or use a spreadsheet then you could construct your own solver

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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