# Using Quadratic formula in other fields

• ramsey2879
In summary, the discovery of the complex number field was a result of resolving the quadratic formula for negative values of b^2 - 4ac. The quadratic formula can be used to solve zM^2 + wM - t = 0 in the complex number field, even when z, w, or t are fractional powers. In the complex numbers, every number has two square roots, with one being the additive inverse of the other. It is important to note that there is no such thing as a "positive" complex number unless it is a real number. For example, the two square roots of 4i2 = -4 are 2i and -2i, and for 4i, they are \
ramsey2879
Since the the discovery of the complex number field resulted in part from efforts to resolve the quadratic formula when b^2 - 4ac was negative, am I right to use the quadratic formula to solve $$zM^2 + wM - t = 0$$ in the complex number field even when z, w or t can be fractional powers?

"Fractional powers" of what? As long as you have chosen a specific branch so the roots are well defined, they are just numbers. Yes, the quadratic formula applies.

Hyper-stupid question: in complex numbers, does the square root have also two values, one plus and one minus? If so, how would the square root of (4 i^2) look like, for example? I ask, of course, because this is present in the usual formula for solving quadratic equations; and all the same if you 'complete the square'.

Yes, in the complex numbers, every number has two square roots, one the additive inverse of the other. I am reluctant to say "one plus and the other minus" since that could be construed to mean "positive and negative" which has no meaning for general complex numbers. There is no such thing as a "positive" complex number unless it happens to be real.

The two square roots of 4i2 = -4 are 2i and -2i, of course. A little more complicated question would be the two square roots of 4i. They are $\sqrt{2}+ i\sqrt{2}$ and $-\sqrt{2}-i\sqrt{2}$.

## 1. What is the quadratic formula and how is it used in other fields?

The quadratic formula is a mathematical formula used to solve quadratic equations, which are equations with a variable raised to the second power. It is written as x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are coefficients in a quadratic equation of the form ax^2 + bx + c = 0. This formula is used in various fields such as physics, engineering, and economics to solve problems involving quadratic equations.

## 2. Can the quadratic formula be used to solve real-world problems?

Yes, the quadratic formula can be used to solve real-world problems in various fields. For example, it can be used in physics to calculate the maximum height of a projectile, in engineering to determine the optimal dimensions of a structure, and in economics to find the break-even point for a business.

## 3. How do you know when to use the quadratic formula in a problem?

You can use the quadratic formula to solve any quadratic equation, regardless of its form. However, it is most useful when the equation cannot be easily factored or when the roots of the equation are irrational or complex numbers. If you are unsure whether to use the quadratic formula, you can always try factoring the equation first.

## 4. Are there any limitations to using the quadratic formula?

There are a few limitations to using the quadratic formula. Firstly, it can only be used to solve quadratic equations, so it is not applicable to other types of equations. Additionally, the equation must be in standard form (ax^2 + bx + c = 0) for the formula to work. Lastly, if the coefficients a, b, and c are very large or very small, the formula may produce inaccurate results due to rounding errors.

## 5. Can the quadratic formula be used to find both real and imaginary solutions?

Yes, the quadratic formula can be used to find both real and imaginary solutions. If the value inside the square root is negative, the equation will have two complex solutions. These solutions can be written in the form a ± bi, where a and b are real numbers and i is the imaginary unit (√-1). Real solutions will have a value of 0 for b, making the second term in the solution disappear.

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