Real Roots of Composite Polynomials: Solving P(Q(x))=0

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In summary, $P(Q(x))=0$ means finding values of $x$ that make the composite function $P(Q(x))$ equal to zero. To solve this, we can use algebraic or numerical methods. Real roots are values of $x$ that make $P(Q(x))$ equal to zero, and there can be more than one real root. The real roots of $P(Q(x))=0$ are the same as the roots of $Q(x)$.
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Let $P(x)=x^2+\dfrac{x}{2}+b$ and $Q(x)=x^2+cx+d$ be two polynomials with real coefficients such that $P(x)Q(x)=Q(P(x))$ for all real $x$. Find all real roots of $P(Q(x))=0$.
 
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Since $P(x)$ divides $Q(P(x))=(P(x))^2+cP(x)+d$, it follows that $P(x)$ divides $d$ and hence $d=0$. Thus $P(x)Q(x)=P(x)^2+cP(x)$ and hence $Q(x)=P(x)+c$. This gives $c=Q(x)-P(x)=\left(c-\dfrac{1}{2}\right)x-b$ and thus $c=\dfrac{1}{2}$ and $b=c=-\dfrac{1}{2}$. Consequently, $P(x)=x^2+\dfrac{x}{2}-\dfrac{1}{2}$ and $Q(x)=x^2+\dfrac{x}{2}$ and

$\begin{align*}P(Q(x))&=\left(x^2+\dfrac{x}{2}\right)^2+\dfrac{1}{2}\left(x^2+\dfrac{x}{2}\right)-\dfrac{1}{2}\\&=\dfrac{1}{4}(4x^4+4x^3+3x^2+x-2)\end{align*}$

Clearly $P(Q(-1))=0$ and any other rational root will be $\pm\dfrac{1}{4},\,\pm \dfrac{1}{2}$. Of these, $\dfrac{1}{2}$ is a root. The remaining factor is $h(x)=x^2+\dfrac{x}{2}+1$ and this has no real roots.
 

What is the meaning of $P(Q(x))=0$?

The equation $P(Q(x))=0$ means that the function $P$ composed with the function $Q$ results in a value of 0. In other words, the input $x$ is being transformed by $Q$, and then the output of $Q$ is being plugged into $P$, resulting in a final output of 0.

How do you solve $P(Q(x))=0$ for real roots?

To solve $P(Q(x))=0$ for real roots, you need to first determine the function $Q(x)$ and then plug it into the function $P(x)$. This will result in a new equation, which can then be solved using various methods such as factoring, the quadratic formula, or graphing.

Why is it important to specify "real roots" when solving $P(Q(x))=0$?

Specifying "real roots" is important because it indicates that we are only interested in solutions that are real numbers. In other words, we are looking for values of $x$ that will make the equation $P(Q(x))=0$ true in the real number system. This is important because some equations may have complex solutions, which are not considered real roots.

Can $P(Q(x))=0$ have more than one real root?

Yes, it is possible for $P(Q(x))=0$ to have more than one real root. This will depend on the specific functions $P(x)$ and $Q(x)$ and how they interact with each other. For example, if $P(x)$ is a quadratic function and $Q(x)$ is a linear function, the resulting equation $P(Q(x))=0$ could have two real roots.

What are some real-life applications of solving $P(Q(x))=0$ for real roots?

The concept of solving $P(Q(x))=0$ for real roots has many real-life applications, such as in physics, engineering, and economics. For example, in physics, this equation can be used to model the motion of a projectile, where $P(x)$ represents the vertical position and $Q(x)$ represents the horizontal position. In engineering, this equation can be used to design circuits, where $P(x)$ represents the voltage and $Q(x)$ represents the current. In economics, this equation can be used to analyze supply and demand, where $P(x)$ represents the price and $Q(x)$ represents the quantity.

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