# Solving Quadratic Equations with a New Method

• abia ubong
In summary, the author claims to have found a method for solving quadratic equations that is more complicated than the existing quadratic formula. However, he provides no evidence to support his claim, and it is likely incorrect.
abia ubong
hey , i tried deriving another method for solving a quadratic equation ,and here is wat i came across
+or - sqrt ([b^2-2ac+or-(b sqrt[b^2-4ac])/2a^2) i hope u get this correctly
its read plus or minus square root of b^2-2ac plus or minus b root b^2-4ac

all divided by 2a^2,remember the first root is common to all functions.i know u will get four values of which 2 will be eligible.pls help me confirm the formula for all eqns.thnxs

I notice nobody has responded yet, maybe they know something I don't. But I'm going to give you the benefit of the doubt once and assume that you are posting the legitimate results of your efforts. If it turns out you're just posting nonsense to spam the forums, then I'm wasting my time. We'll see.

I interpret what you wrote as this:

$$\pm \sqrt{ b^2 - 2ac \pm \frac{b\sqrt{b^2-4ac}}{2a^2}}$$

I'm curious to see how you "derived" it. I have to say I'm very doubtful that this is a correct formula. For two reasons:

1. It's only valid if it is equivalent to the standard quadratic formula, and I couldn't get it to the standard form using algebraic manipulations, not that I tried very hard. Also, if it is correct, ie equivalent to the existing quadratic formula, then it is useless, because it is in a much more complicated form than the standard quadratic formula.

2. A formula for calculating the roots of a quadratic should return only two roots, no more, no less. If the coefficients a, b, and c are real numbers, then those roots should be either real and distinct, or a real double root, or a complex conjugate pair. It makes no sense that 2 legitimate roots and 2 dummy roots would be returned.

In other words, you have some explaining to do... :tongue2:

Edit: Have you seen the derivation of the real quadratic formula, and/or tried it yourself? I think it would be a useful exercise. Start with a quadratic polynomial in the form:

$$ax^2 + bx + c$$

Complete the square, and solve for x.

Last edited:
If your interpretation of the original ASCII version of the new "quadratic equation" is right, then it's wrong by a simple inspection. The simplest quadratic equation

$$x^2+2x+1=0$$

will have solutions of $$\sqrt{2}$$ and $$-\sqrt{2}$$ via his method while we all know the solution is -1 with multiplicity 2, just to change things up a bit

not that cepheid,2a^2 is a denominator to all not just bsqrt...

so pls give it another try houserichichi

Is it too difficult for you to read the responses you get? You have two $\pm$ in you formula. How many solutions to a quadratic equation does that give you? How many solutions does a quadratic equation have?

What solutions does your formula give for x2- 2x+ 1= 0?

You are going to have to do something with those $\pm$s!

Would you mind showing us how you derived that formula.

By the way: 1) It's not a good idea to give your e-mail address out publicly.
2) If you have to call yourself "boygenius" then you aren't!

thnxs hallsofivy ,but for the mail, it was opened at first for me by my brother and i have been using it so u see i can not change it for another or els i would have

## 1. How is this new method different from traditional methods of solving quadratic equations?

This new method involves using a different formula, known as the quadratic formula, to find the solutions to a quadratic equation. This formula can be used for any type of quadratic equation, whereas traditional methods may only work for specific types of equations.

## 2. Is this new method more efficient than traditional methods?

It depends on the specific equation and the individual's proficiency with both methods. In some cases, the new method may be quicker and easier to use, while in others traditional methods may be more efficient.

## 3. Can this new method be applied to all quadratic equations?

Yes, the quadratic formula can be used to solve any quadratic equation, regardless of the coefficients or variables involved. However, it may not always result in real solutions, as some equations may have complex or imaginary solutions.

## 4. Are there any limitations to using this new method?

The main limitation of this method is that it can only be used for quadratic equations. It cannot be applied to equations with higher powers or involving different types of functions.

## 5. How can I check if the solutions obtained using this new method are correct?

You can always plug the solutions back into the original equation to verify if they satisfy the equation. Additionally, you can use a graphing calculator or software to plot the equation and see if the solutions align with the points of intersection.

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