Question

Let $\alpha ,\beta $ be the roots of the equation ${{x}^{2}}-ax+b=0$ and ${{\text{A}}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}.$ Then \[{{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}=\]
$\left( a \right) -a$
$\left( b \right) b$
$\left( c \right) 0$
$\left( d \right) a-b$

Answer 1

Hint: The roots of an equation always satisfy the equation. Both of the given roots $\alpha ,\beta $ always satisfy the given equation. We will apply both of the given roots in the given equation to find the required value.

Complete step by step solution:
Let us suppose that $\alpha $ and $\beta $ are the roots of the equation ${{x}^{2}}-ax+b=0.$
We know that the roots of an equation always satisfy the equation. So, both $\alpha $ and $\beta $ satisfy the given equation ${{x}^{2}}-ax+b=0.$
Let us apply each of these roots in the equation.
When we apply $\alpha $ in the equation ${{x}^{2}}-ax+b=0,$ we will get
$\Rightarrow {{\alpha }^{2}}-a\alpha +b=0.......\left( 1 \right)$
When we apply $\beta $ in the equation ${{x}^{2}}-ax+b=0,$ we will get
$\Rightarrow {{\beta }^{2}}-a\beta +b=0.......\left( 2 \right)$
It is given that ${{\text{A}}_{n}}={{\alpha }^{n}}+{{\beta }^{n}}.$
So, we can say that ${{\text{A}}_{n+1}}={{\alpha }^{n+1}}+{{\beta }^{n+1}}$ and ${{\text{A}}_{n-1}}={{\alpha }^{n-1}}+{{\beta }^{n-1}}.$
Now, let us substitute these in \[{{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}.\]
So, we will get \[{{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}={{\alpha }^{n+1}}+{{\beta }^{n+1}}-a\left( {{\alpha }^{n}}+{{\beta }^{n}} \right)+b\left( {{\alpha }^{n-1}}+{{\beta }^{n-1}} \right).\]
Thus, when we open the brackets we will get,
\[\Rightarrow {{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}={{\alpha }^{n+1}}+{{\beta }^{n+1}}-a{{\alpha }^{n}}-a{{\beta }^{n}}+b{{\alpha }^{n-1}}+b{{\beta }^{n-1}}.\]
In the next step, let us rearrange the equation accordingly to get the following,
\[\Rightarrow {{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}={{\alpha }^{n+1}}-a{{\alpha }^{n}}+b{{\alpha }^{n-1}}+{{\beta }^{n+1}}-a{{\beta }^{n}}+b{{\beta }^{n-1}}.\]
Thus, we will get,
\[\Rightarrow {{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}=\left( {{\alpha }^{n+1}}-a{{\alpha }^{n}}+b{{\alpha }^{n-1}} \right)+\left( {{\beta }^{n+1}}-a{{\beta }^{n}}+b{{\beta }^{n-1}} \right).\]
Let us take the common factors out from both the summands on the left-hand side.
The term ${{\alpha }^{n-1}}$ is the common factor in the first summand and the term ${{\beta }^{n-1}}$ is the common factor in the second summand.
So, we will get
\[\Rightarrow {{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}={{\alpha }^{n-1}}\left( {{\alpha }^{2}}-a\alpha +b \right)+{{\beta }^{n-1}}\left( {{\beta }^{2}}-a\beta +b \right).\]
In the above obtained equation, we can see that the first and second terms include \[{{\alpha }^{2}}-a\alpha +b\] and \[{{\beta }^{2}}-a\beta +b\] respectively.
By applying the equations $\left( 1 \right)$ and $\left( 2 \right)$ in the above equation, it will become
\[\Rightarrow {{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}={{\alpha }^{n-1}}\left( {{\alpha }^{2}}-a\alpha +b \right)+{{\beta }^{n-1}}\left( {{\beta }^{2}}-a\beta +b \right)={{\alpha }^{n-1}}\cdot 0+{{\beta }^{n-1}}\cdot 0=0.\]
Hence \[{{\text{A}}_{n+1}}-a{{\text{A}}_{n}}+b{{\text{A}}_{n-1}}=0.\]

Note: The roots are also called the zeros of the equation. If a value satisfies an equation, then that value is called a root or a zero of the equation. We can use various methods to find the roots of an equation satisfying specific conditions.