Without solving the equation, find the value of the roots

In summary, the problem is to find the value of (\alpha - \beta)^2 given the expressions for \alpha + \beta and \alpha\beta. By using the formula (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta, the solution can be obtained.
  • #1
BOAS
552
19

Homework Statement



23 - 5x - 4x2 = 0

find ([itex]\alpha[/itex] - [itex]\beta[/itex])2



Homework Equations



In previous parts of the question I've calculated [itex]\alpha[/itex] + [itex]\beta[/itex], [itex]\alpha[/itex][itex]\beta[/itex], 1/[itex]\alpha[/itex] + 1/[itex]\beta[/itex] and ([itex]\alpha[/itex]+1)([itex]\beta[/itex]+1) but I can't think of any rules I know to help me solve the problem.

The Attempt at a Solution



Expanding ([itex]\alpha[/itex] - [itex]\beta[/itex])2

gives

[itex]\alpha[/itex]2+[itex]\beta[/itex]2 -2[itex]\alpha[/itex][itex]\beta[/itex]

But I don't know what I can do with this info.

I'd appreciate a nudge in the right direction!

Thanks
 
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  • #2
You can write your expression fairly simply in terms of [itex] \alpha+\beta [/itex] and [itex] \alpha \beta [/itex] (from which you can get the final answer according to your part 2)
 
  • #3
BOAS said:

Homework Statement



23 - 5x - 4x2 = 0

find ([itex]\alpha[/itex] - [itex]\beta[/itex])2



Homework Equations



In previous parts of the question I've calculated [itex]\alpha[/itex] + [itex]\beta[/itex], [itex]\alpha[/itex][itex]\beta[/itex], 1/[itex]\alpha[/itex] + 1/[itex]\beta[/itex] and ([itex]\alpha[/itex]+1)([itex]\beta[/itex]+1) but I can't think of any rules I know to help me solve the problem.

The Attempt at a Solution



Expanding ([itex]\alpha[/itex] - [itex]\beta[/itex])2

gives

[itex]\alpha[/itex]2+[itex]\beta[/itex]2 -2[itex]\alpha[/itex][itex]\beta[/itex]

But I don't know what I can do with this info.

I'd appreciate a nudge in the right direction!

Thanks

##(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta##
You said that you have already calculated ##\alpha + \beta ## and ##\alpha\beta ##.
 
  • #4
Mark44 said:
##(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta##
You said that you have already calculated ##\alpha + \beta ## and ##\alpha\beta ##.

Gah, I should have seen that.

Thanks!
 

Related to Without solving the equation, find the value of the roots

1. What does it mean to find the value of the roots without solving the equation?

Finding the value of the roots without solving the equation means to determine the numerical values of the solutions without explicitly solving for them algebraically. This can be done using various methods such as factoring, graphing, or using the quadratic formula.

2. Can the value of the roots be found without solving the equation?

Yes, the value of the roots can be found without solving the equation. This approach can often be faster and more efficient than solving the equation algebraically.

3. Why would someone want to find the value of the roots without solving the equation?

There are several reasons why someone may want to find the value of the roots without solving the equation. Some possible reasons include having limited time to solve the equation, wanting to avoid complicated algebraic calculations, or wanting to verify the solutions obtained from solving the equation algebraically.

4. What are some techniques for finding the value of the roots without solving the equation?

There are multiple techniques for finding the value of the roots without solving the equation, such as using the sum and product of roots, the rational root theorem, and the discriminant. Other methods include graphing the equation and using the quadratic formula.

5. Are there any limitations to finding the value of the roots without solving the equation?

While finding the value of the roots without solving the equation can be a useful approach, it may not always be possible or accurate. Some equations may not have real solutions, making it impossible to find the roots without solving the equation. Additionally, certain methods may only provide estimates or approximations of the roots, which may not be entirely accurate.

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