Basically how to solve an equation like \(x^6+x^3-6=0\) or anything of the sort. The thing to keep in mind is this: the value of the exponent of the first \(x\) should be twice the value of the exponent of the second \(x\). Think of it like this: \(ax^m+bx^{\frac{m}{2}}+c=0\), or like \(ax^{2m}+bx^m+c=0\).
Let's work with \(x^4-5x^2+4=0\).
The first thing to look at here is the exponents. 2 is half of 4, therefore we can continue.
Now, let's replace \(x^2\) with another variable, \(p\) for example: \(x^2=p\). If this is the case, then \(x^4=p^2\).
Our new equation is \(p^2-5p+4=0\). This is a regular quadratic. Solve this however you want.
The values of \(p\) are 4 and 1. This is not the end!
Recall \(x^2=p\). Put our values of \(p\) into this:
Now we have our values of \(x\).
Let's do another: \(6x+\sqrt{x}-1=0\)
First, we need to recognise that \(\sqrt{x}=x^{\frac{1}{2}}\).
Next, we need to recognise that \(\frac{1}{2}\) is half of 1.
Now, we repeat the same procedure as before. Let's use \(h\): \(x^{\frac{1}{2}}=h\) and \(x=h^2\).
Our equation is \(6h^2+h-1=0\). This gets us \(h=\frac{1}{3}\) and \(h=-\frac{1}{2}\).
Put these values into our substitution:
Note that your answer will be \(\pm\) only if you take a square root.
One more: \(tan^2(x)=1+tan(x)\). Let's rearrange this first: \(tan^2(x)-tan(x)-1=0\).
Something to note is that \(tan^2(x)=(tan(x))^2 \neq tan(x^2)\). Hence, to simplify things a bit: \((tan(x))^2-tan(x)-1=0\).
The rest is the same as before: let \(tan(x)=f\) and \((tan(x))^2=f^2\).
We're left with \(f^2-f-1=0\). Solved, we get \(f=\frac{1+\sqrt{5}}{2}\) and \(f=\frac{1-\sqrt{5}}{2}\). We won't round off just yet.
Next, we input this back into our subsitutions:
I think that's enough.
This article was written on 25/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!