Roots of a quadratic equation.

Question

Assume I have an equation

$$a(f(x))^2+bf(x)+c=0.$$ I am required to find the roots of the equation.

When I solve this equation I will obtain a value of $x$, say $p$ and that value of $x$ will give me a value of $f(x)$, say $q$ (i.e. $f(p)=q$). So my doubt is whether $p$ will be called as the root of the equation or is it $q$ that is the root of the equation?

I know my doubt is very silly but please forgive me. I am not that good at math. Thanks a lot for bearing with me and helping me.

Answer 1

Let's define a polynomial $p(x) = ax^2+bx+c$. You need to solve $$ \tag{1} p(f(x)) = 0. $$

To do that, as you correctly note, you first find the roots of $p(x)$, say $p_+$ and $p_-$, so that $p(p_+) = 0 = p(p_-)$, and then solve $f(x) = p_+$ and $f(x) = p_-$, obtaining some solutions, say $x^*$. These final solutions $x^*$ will be solutions of (1) and roots of $p(f(x))$.

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UPDATE

Not sure how a function can be a root of itself. If you mean $p_-$ or $p_+$, then indeed, they are the roots of $p$, so $p(p_+) = p(p_-) = 0$, but they are not necessarily the roots of $p(f(x))$.

For example, let $p(x) = x^2-16x+64 = (x-8)^2$ and $f(x) = x^3$. Note that:

• $p(x)$ has roots of $p_- = 8$ and $p_+ = 16$
• $f(x)$ itself only has the root at $0$
• $p(f(x))$ has a root at $f(x) = 8 \iff x = 2$ and $f(x) = 16 \iff x = 2\sqrt[3]{2}$

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UPDATE 2

Let's take another particular case and consider $p(x) = (x-1)^2$ and $f(x) = \tan x$. Then, to solve $$\tag{2} (\tan x - 1)^2 = 0$$ you:

• find all roots of $p(x) = (x-1)^2$, which are $x = 1$
• solve $f(x) = 1$, in other words, $\tan x = 1$, so $x_n = \pi/4 +n\pi$ for any integer $n$.

Thus, roots of (2) are the numbers $(x_n)_{n=-\infty}^\infty$.

Answer 2

When you solve the equation, you will get a value for f(x) first. That would be the root: $r=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Then you would solve $f(x)=r$ for x.

Answer 3

You're not going to obtain a value of $x$ when you solve the equation; you'll obtain a value of $f(x)$, as per the quadratic formula:

$$f(x)=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Answer 4

The equation $a(f(x))^2+ b(f(x))+ c= 0$ is "quadratic" only if f(x) is itself a linear function. If u is a solution to $au^2+ bu+ c= 0$ (there might be two values of u) then any solution to f(x)= u is solution to $a(f(x))^2+ b(f(x))+ c= 0$. How many that is depends on the function, f.