# Want to solve a quadratic equation

• whyonlyme
In summary, the conversation revolves around finding the roots of the cubic equation m^3-m^2+2=0. The participants suggest various methods such as substitution, using the factor theorem, and using synthetic division to solve the equation. The conversation also touches upon the importance of checking solutions and the need for basic understanding of solving cubic equations. Overall, the focus is on helping the person asking the question to understand and solve the problem.
whyonlyme

## Homework Statement

Actually, it is told by one of my friend. He asked for the answer of this quadratic equation.

m^3-m^2+2=0

m^3-m^2+2=0

## The Attempt at a Solution

But, I think we need one more equation to solve it. Is there any way to solve it by only one equation? If yes, please describe it or if not, then tell me the proper reason..

That is not a quadratic equation; it's a cubic equation.

Okay.. Sorry then. But, is there any possible solution?

whyonlyme said:
Okay.. Sorry then. But, is there any possible solution?

sub in 0, 1, -1 etc and see if you can find a factor

Well, I don't know how to solve it. Can anyone explain me??

There's also a formula (rather involved).

whyonlyme said:
Well, I don't know how to solve it. Can anyone explain me??

Phospho has already given you some valuable suggestions. Have you tried them?

I said that I didn't understand it. I know how to solve quadratic equation, but don't know how to solve cubic equation. That is what I am saying, can anyone explain it.

While the formula for finding the roots of a cubic function would do the trick, I'd suggest you do it using the Factor Theorem/The Remainder Theorem, and use any method of division to reduce the polynomial.

whyonlyme said:
I said that I didn't understand it. I know how to solve quadratic equation, but don't know how to solve cubic equation. That is what I am saying, can anyone explain it.

If you can find one root, say r, of the equation then m-r is a factor of the polynomial. You should be able to then write it as (m-r)*(quadratic)=0. Then solve the quadratic. But you have to find the one root r, first. There is one. Guess it.

whyonlyme said:
I said that I didn't understand it. I know how to solve quadratic equation, but don't know how to solve cubic equation. That is what I am saying, can anyone explain it.

That is not what he said and not what he meant. Why can't you just substitute m = 0, m = 1 or m = -1 into the equation to see if the equation is satisfied?

Nobody is asking you to solve a cubic equation---although people have pointed out to you that there are formulas to do it (just Google 'cubic equation'). What people are asking you to do is to test some m-values to see if they work.

whyonlyme said:
I said that I didn't understand it. I know how to solve quadratic equation, but don't know how to solve cubic equation. That is what I am saying, can anyone explain it.

The easiest way is to look for a potential linear of the form x - a, where a is a specific number. If you can find a linear factor, the other factor of your cubic will be a quadratic, and you can use the quadratic formula to find its roots.

Here's a link to a web page on polynomial long division - http://www.purplemath.com/modules/polydiv3.htm

For your problem, if there are any linear factors with rational coefficients, they must be of the form (x - 1), (x + 1), (x - 2), or (x + 2).

This is not a calculus problem, so I'm moving it to the precalc section.

I searched for the cubic equation solver online. I found results like this

m1= 2.73205080756888
m2= -0.73205080756888
m3= 1

I saw the method but didn't get it. I have to study it from the basic. btw, thnx for the help..

whyonlyme said:
I searched for the cubic equation solver online. I found results like this

m1= 2.73205080756888
m2= -0.73205080756888
m3= 1

I saw the method but didn't get it. I have to study it from the basic. btw, thnx for the help..

Have you checked to see if these work? All three are WRONG!

On several occasions you have been asked to try some simple values such as m = -1, m = 1, etc., to see if they solve the equation. You have refused to do this, claiming you do not know how to solve cubic equations. As I said already, just checking if a given value "works" is not solving---it is just "checking".

What people were trying to do is to *teach*you how to solve cubic equations, but you have not been cooperative.

OKay... I tried to put m= -1, 1 and 0..

It does not satisfy the equation.

whyonlyme said:
OKay... I tried to put m= -1, 1 and 0..

It does not satisfy the equation.

Try m = -1 again.

whyonlyme said:
OKay... I tried to put m= -1, 1 and 0..

It does not satisfy the equation.

That is not true. Try again. Perhaps you are looking at a different equation? Are you sure your equation is ##m^3 - m^2 + 2 = 0?##

Last edited:
whyonlyme said:
OKay... I tried to put m= -1, 1 and 0..

It does not satisfy the equation.

Oh come on, m=(-1) does.

whyonlyme said:
OKay... I tried to put m= -1, 1 and 0..
It does not satisfy the equation.
Don't try to use all three numbers as the three roots to the equation, just try the -1 as one of the roots.

The equation m3-m2+2=0 can be written also as (m3+1)+(1-m2)=0

You can factorise both terms inside the parentheses, and there is a common factor.

ehild

Try synthetic division.

## What is a quadratic equation?

A quadratic equation is a mathematical expression that contains a variable raised to the second power, such as x^2, and can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants.

## How do you solve a quadratic equation?

To solve a quadratic equation, you can use the quadratic formula, which is x = (-b ± √(b^2-4ac))/2a. You can also use factoring, completing the square, or graphing to find the solutions.

## Why do we need to solve quadratic equations?

Quadratic equations are used to solve real-world problems, such as finding the maximum or minimum value of a quadratic function, determining the path of a projectile, or finding the dimensions of a rectangle with a given area.

## What are the different types of solutions for a quadratic equation?

A quadratic equation can have two distinct real solutions, one repeated real solution, or two complex solutions. The type of solutions depends on the discriminant, b^2-4ac, of the equation.

## How can I check if my solution to a quadratic equation is correct?

You can check your solution by plugging it back into the original equation and seeing if it satisfies the equation. You can also use a graphing calculator to graph the equation and check if your solution is the x-intercept.

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