# Is the use of the quadratic formula valid? (Flammable Math's "How REAL Men Solve Equations" video)

I was wondering whether the steps used in Flammable Math's YouTube video "How REAL Men Solve Equations" are valid or not. It's a joke video, but all of his work seems correct. Is this a valid use of the quadratic formula?

$$8x=1\Rightarrow 4x+4x-1=0$$ $$\Rightarrow (x)2^2+(2x)2-1=0$$ $$\Rightarrow 2_{12} = \frac{-2x\pm \sqrt{4x^2+4x}}{2x}$$ $$\require{cancel} = \frac{-\cancel{2x}\pm\cancel{2x} \sqrt{1+\frac{1}{x}}}{\cancel{2x}}$$ $$\stackrel{2>0}{\Rightarrow} 2 = -1+\sqrt{1+\frac{1}{x}}\Rightarrow 3 = \sqrt{1+\frac{1}{x}}$$ $$\Rightarrow 9 = 1+\frac{1}{x} \Rightarrow \frac{1}{8}=x$$

I believe it is because if you let $$2 = y$$, you get a quadratic equation in terms of $$y$$, then you can set up the quadratic formula to solve for $$y$$.

Like this $$(x)(y)^2+(2x)y-1 = 0$$. So, we get $$y = \frac{-2x\pm \sqrt{4x^2+4x}}{2x}$$. Now that renaming made the manipulation make much more sense.

• Remember kids: work harder, not smarter. Dec 31, 2020 at 2:49
• It'd be better if you transcribed the equations here - we don't like questions to rely on images. And is there any bit in particular you are wondering about. (It looks fine to me and you seem to think it looks fine, which leaves little to ask about!) Dec 31, 2020 at 3:11
• Silly though it may be, there is nothing wrong with this: there is nothing 'magical' about variables as opposed to numbers, and the same rules apply to each. The quadratic formula is just the result of some basic algebraic manipulations (specifically, the process of completing the square); we can perform those with numbers, variables, or a mixture of the two. Dec 31, 2020 at 3:11
• Thank you, Noah, that answers my question Dec 31, 2020 at 3:14
• There are some (extremely contrived) problems where this is actually a reasonable method of solution. For example, the equation $5-x^2=\sqrt{5-x}$ can be solved more easily by treating it as a quadratic in $5$ than as a quartic in $x$. Dec 31, 2020 at 4:09

A common error in mathematicians is thinking that particular letters or symbols carry any significance. The most common statement of the quadratic formula is that given real numbers $$a,b,c$$, the solutions of the polynomial equation $$ax^2+bx+c=0$$ Are $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ But one could just as easily say that given real numbers $$\#,\%,\&$$ the solutions of the polynomial equation $$\#@^2+\%@+\&=\text{zero}$$ Are $$@=\frac{\text{negative }\% ~\mp \square \text{root}\{{}^2\% ~\_~ \text{IV}\#\&\}}{2\#}$$ I'll leave it to you to work out what the odd symbols mean. The quadratic formula (and mathematics in general) is (are) a statement(s) about numbers (or other mathematical objects), not symbols.

Let's suppose we have the equation $$4x+3=11$$. We can solve this using elementary methods to obtain $$x=2$$: $$4x=11-3$$ $$x=\frac{11-3}{4}=8/4=2$$ Let's start again. Is it incorrect to now say $$4\cdot 2+3=11$$ ? No. Is it incorrect to say $$4\cdot 2=11-3$$? No. Is it incorrect to say $$2=\frac{11-3}{4}$$ ? Finally, is it incorrect to say $$2=2$$? Erm.. no. When we say "$$x$$ equals $$2$$ is a solution of the equation $$4x+3=11$$" precisely what we mean is that if we substitute the numeric value $$2$$ in place of $$x$$ in the equation, that the numeric value of both sides of the equation are the same (otherwise I suppose it wouldn't be an equation).

Another edit:

To reaffirm that the quadratic formula is a statement about numbers, and not symbols, I will present the quadratic formula without using symbols. Here goes:

Imagine any three numbers - call them the first, second and third. The only demand is that the first cannot be zero. If we can find a number, call it the 'square root' such that when we multiply it by itself we get the square of the second minus four times the product of the first and the third, then we can call half the 'square root' divided by the first, minus half the second divided by the first, the 'solution', and this 'solution' has the remarkable property that if we square it, multiply that result by the first, then add the product of the second and "solution" and then add the third, we will get zero, no matter what our choices for the first, second, and third are.

I think this makes it clear why we use symbols rather than words in mathematics!

• The odd symbols each represent a symbol in the original quadratic equation. However, I do not understand why you did that? Are you trying to show me that numbers and variables don't really matter and that you can substitute them in for each other? Dec 31, 2020 at 3:27
• @SomeGuy See my edit. Dec 31, 2020 at 3:37
• Oh, I see, we only call a number x because we don't know the numerical value, so we just call it a letter. However, if we do know the solution, we can just let x = the answer and still solve it right. Dec 31, 2020 at 3:41
• Exactly! Ultimately most things in mathematics boil down to numbers, but sometimes using letters is so much easier. Dec 31, 2020 at 3:51

Example 1 :

The method can be used for the following equation : $$\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}+bdx^{2n}+cdx^n+d^2=0\tag1$$ where $$n$$ is a positive integer, and $$a,b,c,d$$ are real numbers with $$c\not=0$$.

Since $$(1)$$ can be written as $$d^2+d(bx^{2n}+cx^n)+\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}=0$$ which can be seen as a quadratic equation in $$d$$, we see that the discriminant is \begin{align}&(bx^{2n}+cx^n)^2-4\bigg(\frac{abc-a^2}{c^2}x^{4n}+ax^{3n}\bigg) \\\\&=\bigg(\frac{bc-2a}{c}\bigg)^2x^{4n}+2(bc-2a)x^{3n}+c^2x^{2n} \\\\&=\bigg(\dfrac{bc-2a}{c}x^{2n}+cx^n\bigg)^2\end{align} from which we obtain $$d=\frac{-bx^{2n}-cx^n\pm\bigg(\dfrac{bc-2a}{c}x^{2n}+cx^n\bigg)}{2}$$ which are quadratic equations in $$x^n$$.

Example 2 :

This is an expansion of @Micah's comment.

The method can be used for the following equation : $$\sqrt{ax^n+b}\ =-(a-c)^2x^{2n}+cx^n+b\tag2$$ where $$n$$ is a positive integer, and $$a,b,c$$ are real numbers.

(Micah's example $$5-x^2=\sqrt{5-x}\$$ is the case where $$(n,a,b,c)=(1,-1,5,0)$$.)

Squaring the both sides of $$(2)$$ gives $$b^2-\bigg(2(a-c)^2x^{2n}-2cx^n+1\bigg)b+(a-c)^4x^{4n}-2c(a-c)^2x^{3n}+c^2x^{2n}-ax^n=0$$ which can be seen as a quadratic equation in $$b$$.

The discriminant is \begin{align}&\bigg(2(a-c)^2x^{2n}-2cx^n+1\bigg)^2-4\bigg((a-c)^4x^{4n}-2c(a-c)^2x^{3n}+c^2x^{2n}-ax^n\bigg) \\\\&=4(a-c)^2x^{2n}+4(a-c)x^n+1 \\\\&=(2(a-c)x^n+1)^2\end{align} so we get $$b=\frac{2(a-c)^2x^{2n}-2cx^n+1\pm(2ax^n-2cx^n+1)}{2}$$ which are quadratic equations in $$x^n$$.

Example 3 :

The method can be used for the following equation : $$\sqrt{ax^n+b}\ +\sqrt{\frac{(2da-a+c)^2}{2d}x^{2n}+cx^n+d^2-2bd+b}\ =d\tag3$$ where $$n$$ is a positive integer, and $$a,b,c,d$$ are real numbers with $$d\gt 0$$.

Squaring the both sides of $$\sqrt{\frac{(2ad-a+c)^2}{2d}x^{2n}+cx^n+b+d^2-2bd}\ =d-\sqrt{ax^n+b}$$ gives $$\frac{(2ad-a+c)^2}{2d}x^{2n}+cx^n+b+d^2-2bd=d^2-2d\sqrt{ax^n+b}\ +ax^n+b$$ which can be simplified as $$\sqrt{ax^n+b}\ =-\bigg(a-\frac{a-c}{2d}\bigg)^2x^{2n}+\frac{a-c}{2d}x^n+b$$

This is of the form $$(2)$$, so treating this as a quadratic equation in $$b$$ works.

If we want to solve $$4x^{20}-4x^{15}-15x^{10}+3x^5+9=0$$ which is of the form $$(1)$$, then we have \begin{align}&3^2+3(-5x^{10}+x^5)+4x^{20}-4x^{15}=0 \\\\&\iff 3=\frac{5x^{10}-x^5\pm \sqrt{(3x^{10}+x^5)^2}}{2} \\\\&\iff 3=4x^{10}\qquad \text{or}\qquad 3=x^{10}-x^5 \\\\&\iff x^{10}=\frac 34\qquad\text{or}\qquad x^5=\frac{1\pm\sqrt{13}}{2} \\\\&\iff x=\pm\sqrt[10]{\frac 34},\ \sqrt[5]{\frac{1+\sqrt{13}}{2}},\ -\sqrt[5]{\frac{\sqrt{13}-1}{2}}\end{align}

If we want to solve $$\sqrt{2x^4+7}\ =-x^{8}+x^4+7$$ which is of the form $$(2)$$, then squaring the both sides, we get \begin{align}&7^2+7(-2x^8+2x^4-1)+x^{16}-2x^{12}+x^8-2x^4=0 \\\\&\implies 7=\frac{2x^8-2x^4+1\pm\sqrt{(2x^4+1)^2}}{2} \\\\&\implies 7=x^8+1\qquad\text{or}\qquad 7=x^8-2x^4 \\\\&\implies x^4=\pm\sqrt 6,\ 1\pm 2\sqrt{2}\end{align} It follows from $$x^4\in\bigg[0,\dfrac{1+\sqrt{29}}{2}\bigg)$$ that the solutions are $$x=\pm \sqrt[8]6$$

If we want to solve $$\sqrt{20x^{6}+x^3-14}\ +\sqrt{x^3+6}\ =10$$ which is of the form $$(3)$$, then squaring the both sides of $$\sqrt{20x^{6}+x^3-14}\ =10-\sqrt{x^3+6}$$ gives $$\sqrt{x^3+6}\ =-x^6+6$$ which is of the form $$(2)$$. Squaring the both sides, we obtain \begin{align}&6^2+6(-2x^6-1)+x^{12}-x^3=0 \\\\&\implies 6=\frac{2x^6+1\pm\sqrt{(2x^3+1)^2}}{2} \\\\&\implies 6=x^6+x^3+1\qquad\text{or}\qquad 6=x^6-x^3 \\\\&\implies x^3=\frac{-1\pm\sqrt{21}}{2},\ \frac{1\pm 5}{2}\end{align}

It follows from $$x^3\in \bigg[-\sqrt 6,\dfrac{-1-\sqrt{1121}}{40}\bigg]\cup\bigg[\dfrac{-1+\sqrt{1121}}{40},\sqrt 6\bigg]$$ that the solutions are $$x=-\sqrt[3]2,\ \ \sqrt[3]{\frac{-1+\sqrt{21}}{2}}$$

• What exactly are you trying to show here? Jul 6, 2021 at 0:01