Roots of a quadratic equation

In summary, the conversation discusses solving a quadratic equation using factorization and the use of a new method involving the relationship between coefficients and roots. However, it is determined that the use of this method is not necessary and completing the square is a more efficient method for solving the equation. The conversation also mentions the idea of finding an equation with roots that are the inverse of a given equation.
  • #1
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Solve the following quadratic equation. Use factorisation if possible.

X2 - 4X - 8 = 0


Normally I wouldn't have trouble factorising a quadratic, but I have just been introduced to a new way to do it and I want to use this way to answer the question.

Here's how far I get, then I'm unsure what to do with the info I have.
X2 - 4X - 8 = 0

Let [itex]\alpha[/itex] and [itex]\beta[/itex] be the roots of the equation.

(X-[itex]\alpha[/itex])(X-[itex]\beta[/itex]) = 0

Therefore

X2 - ([itex]\alpha[/itex]+[itex]\beta[/itex])X + [itex]\alpha[/itex][itex]\beta[/itex] = 0

Or in another form;

X2 + [itex]\frac{b}{a}[/itex]X + [itex]\frac{c}{a}[/itex] = 0

Comparing coefficients.

[itex]\alpha[/itex]+[itex]\beta[/itex] = -[itex]\frac{b}{a}[/itex] = 4

[itex]\alpha[/itex][itex]\beta[/itex] = [itex]\frac{c}{a}[/itex] = -8

And now I'm confused about what I can do with this info to find the factors of the original quadratic.

Any help is appreciated!

Thanks.
 
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  • #2
Hi BOAS! :smile:
BOAS said:
[itex]\alpha[/itex]+[itex]\beta[/itex] = -[itex]\frac{b}{a}[/itex] = 4

So ##\alpha## = 2 + C, ##\beta## = 2 - C. :wink:

(or you could simply complete the square)
 
  • #3
tiny-tim said:
Hi BOAS! :smile:


So ##\alpha## = 2 + C, ##\beta## = 2 - C. :wink:

(or you could simply complete the square)

I don't follow how you got there, please could you elaborate?

I tried completing the square to see if it would make anything clear to me, but I get x = ±√12 - 2 which doesn't seem right :confused:
 
  • #4
Hi BOAS! :smile:
BOAS said:
I don't follow how you got there, please could you elaborate?

If you define C by ##\alpha## = 2 + C, then ##\beta## = … ? :wink:
I tried completing the square to see if it would make anything clear to me, but I get x = ±√12 - 2 which doesn't seem right :confused:

you should have got + 2 :redface:
 
  • #5
tiny-tim said:
Hi BOAS! :smile:


If you define C by ##\alpha## = 2 + C, then ##\beta## = … ? :wink:
:

Where does that definition come from?

you should have got + 2 :redface

Oops - careless mistake.
 
  • #6
BOAS said:
Where does that definition come from?

Isn't it obvious?

If A + B = N, then A and B must be equidistant from N/2 ?
 
  • #7
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  • #8
If you have been given a new method for solving use that. But it can't be what you started doing, i.e. to use these relations between roots and coefficients. Which maybe are what is new to you.

The reason is these relations are symmetric. E.g. in your equations there is nothing to distinguish α from β. (Your expressions are not changed by interchanging α and β.) So you might find a method to eliminate β and end up with an equation with α only. But because youve started with completely symmetrical expressions the same method has got to give you an equation for β - in fact it will be the same equation. You have only changed the name you give to 'a root'. α or β or x. You will just get your starting equation. Just in case you've tried and notice it is like trying to pin down a globule of mercury, this is why. :smile:

In the case you have extra information, if e.g you are told one root is twice the other or other relation between roots, then you can use these relations in the solution - or rather the reduction of degree. But not in the general case.

Added: Posts crossed - you did what I mentioned but if you check out a mistake you made in calculation in the last post you will get an equation in β that is only the starting equation! :tongue:
 
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  • #9
epenguin said:
If you have been given a new method for solving use that. But it can't be what you started doing, i.e. to use these relations between roots and coefficients. Which maybe are what is new to you.

You're correct, I haven't been given a new method for solving equations - The relationship between coefficients is what's new to me.

The reason is these relations are symmetric. E.g. in your equations there is nothing to distinguish α from β. (Your expressions are not changed by interchanging α and β.) So you might find a method to eliminate β and end up with an equation with α only. But because youve started with completely symmetrical expressions the same method has got to give you an equation for β - in fact it will be the same equation. You have only changed the name you give to 'a root'. α or β or x. You will just get your starting equation. Just in case you've tried and notice it is like trying to pin down a globule of mercury, this is why. :smile:

In the case you have extra information, if e.g you are told one root is twice the other or other relation between roots, then you can use these relations in the solution - or rather the reduction of degree. But not in the general case

It sounds like i'd be much better off simply completing the square and using this method for questions that actually require it, such as "find the equation whose roots are the inverse of ..."
 
  • #10
BOAS said:
It sounds like i'd be much better off simply completing the square and using this method for questions that actually require it, such as "find the equation whose roots are the inverse of ..."

Right. But it was an idea, and most people who are thinking actively and meet these relations try a bit to go down that road, which unfortunately leads back to the starting point! :biggrin:
 
  • #11
epenguin said:
Right. But it was an idea, and most people who are thinking actively and meet these relations try a bit to go down that road, which unfortunately leads back to the starting point! :biggrin:

Trial and error :)
 

1. What is a quadratic equation?

A quadratic equation is a mathematical equation that involves a variable raised to the second power, also known as a squared term. Its general form is ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

2. What are the roots of a quadratic equation?

The roots of a quadratic equation are the values of x that make the equation equal to 0. They are also known as the solutions of the equation.

3. How can I find the roots of a quadratic equation?

There are several methods to find the roots of a quadratic equation, such as factoring, completing the square, and using the quadratic formula. The most commonly used method is the quadratic formula, which is x = (-b ± √(b^2 - 4ac)) / 2a.

4. Can a quadratic equation have no real roots?

Yes, a quadratic equation can have no real roots. This happens when the discriminant (b^2 - 4ac) is negative, meaning that the roots would involve imaginary numbers. In this case, the graph of the equation does not intersect with the x-axis.

5. What is the relationship between the roots and the coefficients of a quadratic equation?

The sum of the roots of a quadratic equation is equal to -b/a, and the product of the roots is equal to c/a. In other words, the sum of the roots is the negative of the coefficient of the x-term divided by the coefficient of the squared term, and the product of the roots is equal to the constant term divided by the coefficient of the squared term.

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