Question

If $\alpha $ and $\beta $ are the roots of the equation ${{x}^{2}}+px+q=0$ and ${{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0$, where $n$ is an even integer. Prove that $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ are the roots of the equation ${{x}^{n}}+1+{{\left( x+1 \right)}^{n}}$ .

Answer 1

Hint: To solve this question we need to have the knowledge of polynomials with $''n''$ degree of polynomials. To solve this question we will write the coefficients of the quadratic equation given as the sum of the roots and the products of the root. The next step will be to substitute the roots $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ in place of $x$ in the equations given ${{x}^{n}}+1+{{\left( x+1 \right)}^{n}}$ and these are the roots of the equation or not.

Complete step by step answer:
The question ask us to prove that $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ are the roots of the equation ${{x}^{n}}+1+{{\left( x+1 \right)}^{n}}$ , where $n$ is an even integer when $\alpha $ and $\beta $ are given as the roots of the equation ${{x}^{2}}+px+q=0$ and ${{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0$. The first step is to write the coefficient of the quadratic equation ${{x}^{2}}+px+q=0$ given in terms of $\alpha $ and $\beta $. We know that the sum of the roots is equal to the negative coefficient of $x$. So on doing we get:
$\Rightarrow \alpha +\beta =-p$
The product of the roots is equal to the constant present in the quadratic equation. This will be written as:
$\Rightarrow \alpha \beta =q$
Since, $\alpha $ and $\beta $ are the roots of ${{x}^{2n}}+{{p}^{n}}{{x}^{n}}+{{q}^{n}}=0$, then the equation we will get will be:
$\Rightarrow {{\alpha }^{2n}}+{{p}^{n}}{{\alpha }^{n}}+{{q}^{n}}=0$
$\Rightarrow {{\beta }^{2n}}+{{p}^{n}}{{\beta }^{n}}+{{q}^{n}}=0$
Now we will subtract the above two equation, so as to get the relation between $\alpha $, $\beta $ and
$p$ . On doing this we get:
$\Rightarrow {{\alpha }^{2n}}-{{\beta }^{2n}}+{{p}^{n}}({{\alpha }^{n}}-{{\beta }^{n}})+{{q}^{n}}-{{q}^{n}}=0$
$\Rightarrow ({{\alpha }^{n}}-{{\beta }^{n}})({{\alpha }^{n}}+{{\beta }^{n}})+{{p}^{n}}({{\alpha }^{n}}-{{\beta }^{n}})=0$
$\Rightarrow {{\alpha }^{n}}+{{\beta }^{n}}=-{{p}^{n}}$
The next step is to put $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ in the place of $x$ in the equation ${{x}^{n}}+1+{{\left( x+1 \right)}^{n}}$, and check whether the above equation turns to zero or not. On putting the roots in the equation we get:
$\Rightarrow {{\left( \dfrac{\alpha }{\beta } \right)}^{n}}+1+{{\left( \dfrac{\alpha }{\beta }+1 \right)}^{n}}$
$\Rightarrow {{\alpha }^{n}}+{{\beta }^{n}}+{{\left( \alpha +\beta \right)}^{n}}$
Now we will put the other root, $\dfrac{\beta }{\alpha }$ in the equation. On doing this we will get:
$\Rightarrow {{\left( \dfrac{\beta }{\alpha } \right)}^{n}}+1+{{\left( \dfrac{\beta }{\alpha }+1 \right)}^{n}}$
$\Rightarrow {{\beta }^{n}}+{{\alpha }^{n}}+{{\left( \beta +\alpha \right)}^{n}}$
Now on substituting the relation which are $\alpha +\beta =-p$ and ${{\alpha }^{n}}+{{\beta }^{n}}=-{{p}^{n}}$ in the above equations we will get:
$\Rightarrow {{\beta }^{n}}+{{\alpha }^{n}}+{{\left( \beta +\alpha \right)}^{n}}$
\[\Rightarrow -{{p}^{n}}+{{\left( -p \right)}^{n}}\]
Since negative integer having power even gives positive number, so the above expression will turn to:
\[\Rightarrow -{{p}^{n}}+{{p}^{n}}\]
\[\Rightarrow 0\]
Since the equation ${{x}^{n}}+1+{{\left( x+1 \right)}^{n}}$ turn to zero, so $\dfrac{\alpha }{\beta }$ and $\dfrac{\beta }{\alpha }$ are the roots of the equation.
Hence proved.

Note: While solving these kind of problem do remember that when roots are substitute in the equation the equation should turn to zero. The relation between roots and coefficients are important so as to solve for the known. Some of the relations for the quadratic equation are, the sum of the two roots is negative of the coefficient of $x$, while the product of the roots is equal to the constant of the quadratic equation.