SOLVE INEQUALITIES WITH STEP

     

I just saw this old Vihart clip that mentioned the fact that this polynomial only has an approximate solution, not an exact one. However, when I rearrange the equation into x4 - 1 = -1/x and graph the two individual functions (you can use this online graphing calculator to lớn see), the "approximate" solution of -1.167... Is where these two functions cross. Since they are both continuous, differentiable, & cross at this point, why is this only an approximate solution?


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This is actually a fascinating question, with an even more fascinating answer. First off, it is important to note that there is a real solution khổng lồ that equation. There are also four other complex solutions that you would expect by the fundamental theorem of algebra.

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The real question here is, why can you only approximate the solution. First, off, let's note what tools we have khổng lồ work with. We start with the rational numbers (fractions). The operations we can use are +,-,*,/, & radicals. But since we have all the rational numbers lớn use, we can ignore - & /.

Note that for linear equations, we don't even need radicals to solve them. We just need multiplication and addition. Let's look at a quadratic, x2 - 4x + 2. This has roots 2 + sqrt(2), and 2 -sqrt(2). (Note the symmetry between the two). We can find the root easily with the quadratic formula. Isn't an easy formula for the cubic, but there is a straightforward process that always works. There is also one for quartics (degree 4). But once you get khổng lồ degree 5, there are only special cases where you can express an exact solution using radicals. The reason why you can't is actually very elegant & brings together many different areas of math. It's called Galois Theory.

I'll try to convince you that the roots of some polynomials are fundamentally different. So if you look at the quadratic example above, you can notice that the only irrational number we used was 21/2. If we augment the rationals with this number, we create a new set of numbers where we can solve that polynomial. In fact, this can be done for any polynomial by simply augmenting the rationals with certain numbers. It turns out that these augmented sets are something called a field.

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Lets take another example, x4 - 2 =0. Well, x2 = 21/2, so the roots are +/- 21/4 và +/- i 21/4. If we add i and 21/4 to the rationals, we can make another field. Now picture these roots on the complex plane. The 4 roots lie on the x và y axes, like a baseball diamond.

I hope someone gets lớn this because here's the coolest part. Imagine 4 people running around in a baseball diamond. Suppose you want lớn advance all the runners, just multiply all the roots by i. Let's say you want to lớn swap home plate và second base, take the complex conjugate. You can also imagine a physical square that you can rotate and turn upside down. You can move the vertices but only in certain ways. What if you wanted to swap 1st and 2nd base và leave everything else the same? Well you're out of luck. There's no convenient operation that lets us bởi that. To lớn contrast, you can have a polynomial (x4 - 10x2 +1) with roots sqrt(2) +/- sqrt(3) & -sqrt(2) +/- sqrt(3). In this case the only possible permutations are swapping any two pairs of roots. You can't vị a rotation.

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So, how does this all relate lớn there not being a general solution lớn quintics? Well you can see that radicals can be organized in certain ways, and there will be symmetries that arise from how they are arranged. You can nest radicals inside each other in very complicated ways, but there are limited number of symmetry groups that can arise from this. In order lớn generally solve a degree n polynomial, you have khổng lồ allow for all possible symmetries on n roots. This is called the symmetric group S*n. It is very large and contains n! elements. Once you get to lớn S5. It is "too big. " Radicals alone cannot express the full range of S5*. I hope that gives you some intuition as to lớn what is going on.