# Reducing an Equation to a Quadratic

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:

• $$x^2=4$$
• $$x=\pm2$$
• $$x^2=1$$
• $$x=\pm1$$

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:

• $$x^{\frac{1}{2}}=\frac{1}{3}$$
• $$\sqrt[0.5]{x^{\frac{1}{2}}}=\sqrt[0.5]{\frac{1}{3}}$$
• $$x=\frac{1}{9}$$
• $$x^{\frac{1}{2}}=-\frac{1}{2}$$
• $$\sqrt[0.5]{x^{\frac{1}{2}}}=\sqrt[0.5]{-\frac{1}{2}}$$
• $$x=\frac{1}{4}$$

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:

• $$tan(x)=\frac{1+\sqrt{5}}{2}$$
• $$x=tan^{-1}(\frac{1+\sqrt{5}}{2})$$
• $$x=58.2825... \approx 58.3$$
• $$tan(x)=\frac{1-\sqrt{5}}{2}$$
• $$x=tan^{-1}(\frac{1-\sqrt{5}}{2})$$
• $$x=31.7174... \approx 31.7$$

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!