what is i in factoring equations and how do i understand and apply it?
I am currently in Advanced Algebra-ii and we are working with polynomials however I did not understand the lesson where we learned about the number i or how to apply it when factoring or, in general, using equations. so, my question today is, how does i work? What does it mean? how do I use it in an equation or while factoring?
$\endgroup$ 22 Answers
$\begingroup$The letter $i$ represents $\sqrt{-1}$. The formal definition is $i^2 = -1$. It is used in factoring to factor a sum of squares. The formula is:
$a^2 + b^2 = (a + bi)(a - bi)$
To prove it, I will use LHS-RHS.
LHS = $a^2 + b^2$, RHS = $(a+bi)(a-bi)$
$(a+bi)(a-bi)=a^2+abi-bia-bi^2$
$(a+bi)(a-bi)=a^2-bi^2$
Because $i^2 = -1$, we can simplify the expression a little further
$(a+bi)(a-bi)=a^2-b(-1)$
$(a+bi)(a-bi)=a^2+b^2$
LHS = RHS
There are also many other things you can do with $i$. The number $i$ is actually a complex number, which is also known as an imaginary number. Have you ever tried to solve a quadratic that does not touch the x-axis at all? You would get a negative square root, because $b^2 - 4ac$, the discriminant, is less than $0$. But with the other part of the entire number system, the imaginary numbers, you can find the complex roots. Have you ever wondered why the fundamental theorem of algebra is not always "true"? Just in case you are wondering, the fundamental theorem of algebra is this:
Every polynomial equation that can be written in the form $a_0x^n + a_1x^{n-1} + a_2x^{n-2} + ... + a_{n-2}x^2 + a_{n-1}x^1 + a_nx^0 = 0$ has $n$ roots.
So, for example, every equation of the form $ax^2 + bx + c = 0$ must have $2$ roots, no exceptions. "But sometimes there are no roots", you say. If there are no real roots, then there will be $2$ complex roots.
For example, the equation $x^2 - 3x + 4$ has no real roots. The quadratic formula gives us the equation $x = \frac{10 ± \sqrt{-7}}{2}.$How would we simplify the $\sqrt{-7}$? Just recall that $\sqrt{xy} = \sqrt{x} * \sqrt{y}$. So $\sqrt{-7} = \sqrt{7} * \sqrt{-1}$, which equals $\sqrt{7}i$. So the two roots of $x^2 - 3x + 4$ are $\frac{10 ± \sqrt{7}i}{2}$, which can be simplified to $5 ± \frac{\sqrt{7}i}{2}$.
There is actually a formula for figuring out the complex roots that is very similar to the quadratic formula we are familiar with today, with one slight alteration. The formula is:
$\frac{-b±i\sqrt{b^2 - 4ac}}{2a}$
Where would I graph this, you ask? This brings up to an entirely new coordinate system, the complex plane.
To visually graph and identify the complex roots, we have to work in $3$-D coordinates. A new $i$ axis has to be added. Imagine the $y$ axis actually sticking up toward the "sky", and the $x$ and the $i$ axis traveling on the "floor" in different directions. These all intersect at the origin $(0,0,0)$. We would have to imagine that there is a plane running along the $x$ and $y$ axes, and graph $y = x^2-3x+4$ on that plane. To see the complex roots, we must "flip" the parabola so that it opens downward, and rotate it $90$ degrees so that the parabola is "facing" the $i$ axis to see the complex roots. I cannot explain this well without a visual, so just do a simple Google search on this topic. Here is a direct link to a good resource that talks about everything I mentioned here:
I hope this answer helped you a bit :)
$\endgroup$ 1 $\begingroup$$i=\sqrt{-1}$. The precise definition is $i^2 = -1$. When you allow complex numbers - that is, numbers of the form $a+ib$ where $a,b$ are real numbers, any polynomial of order $n$ with complex coefficients has $n$ roots over the complex numbers (this is the fundamental theorem of algebra). The complex numbers are the algebraic closure of the real numbers (here is the definition of algebraically closed and some examples).
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