If one of the roots of the polynomial $a{{x}^{4}}+b{{x}^{3}}+c{{x}^{2}}+dx+e=0$, where a, b, c, d and e are rational numbers is $\sqrt{2}+\sqrt{3}$ then the other real roots are:
$\begin{align}
& a)\sqrt{2}-\sqrt{3},2,5 \\
& b)\sqrt{2}-\sqrt{3},-2,5 \\
& c)\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},5 \\
& d)\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3} \\
\end{align}$
Answer
599.1k+ views
Hint: Now we are given a fourth degree polynomial and a root of the polynomial. Now we know the relation between the coefficients and roots of the polynomial. Hence using this relation we will find the possibilities of other roots of the polynomial.
Complete step by step answer:
Now a polynomial is an expression in variables and constants.
Here the given polynomial is a fourth degree expression in one variable x.
Now we know that if the degree of polynomial is n then there are n roots of the polynomial.
Hence we know that the given polynomial has 4 roots.
Now in general if polynomial has roots $\alpha ,\beta ,\gamma ,\partial $
Then the polynomial is equal to $\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)\left( x-\partial \right)$
Now the general form of the polynomial in degree 4 is \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{4}}+dx+e\]
Hence on equating we have \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{4}}+dx+e=\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)\left( x-\partial \right)\]
Now on simplifying the RHS of the equation and comparing the coefficients we get standard relation between roots and the coefficients of the polynomial.
Now from the relation we have the sum of the roots must be
$\alpha +\beta +\gamma +\partial =\dfrac{-b}{a}$
Since the coefficients are rational we have $\dfrac{-b}{a}$ is rational.
Hence the sum of the roots must also be rational.
Now let us check the options
If we choose option d then the roots of the expression are $\sqrt{2}+\sqrt{3},\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3}$
In this case we will get the sum of the roots as 0 and hence rational.
So, the correct answer is “Option d”.
Note: Now from the relation between the coefficient and the roots we also have $\alpha \beta +\beta \gamma +\gamma \partial +\partial \alpha +\beta \partial +\alpha \gamma =\dfrac{c}{a}$, $\alpha \beta \gamma +\beta \gamma \partial +\gamma \partial \alpha =\dfrac{-d}{a}$ and the product of the roots which is $\alpha \beta \gamma \partial $ is $\dfrac{e}{a}$. We have the result generalized for polynomials of degree n.
Complete step by step answer:
Now a polynomial is an expression in variables and constants.
Here the given polynomial is a fourth degree expression in one variable x.
Now we know that if the degree of polynomial is n then there are n roots of the polynomial.
Hence we know that the given polynomial has 4 roots.
Now in general if polynomial has roots $\alpha ,\beta ,\gamma ,\partial $
Then the polynomial is equal to $\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)\left( x-\partial \right)$
Now the general form of the polynomial in degree 4 is \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{4}}+dx+e\]
Hence on equating we have \[a{{x}^{4}}+b{{x}^{3}}+c{{x}^{4}}+dx+e=\left( x-\alpha \right)\left( x-\beta \right)\left( x-\gamma \right)\left( x-\partial \right)\]
Now on simplifying the RHS of the equation and comparing the coefficients we get standard relation between roots and the coefficients of the polynomial.
Now from the relation we have the sum of the roots must be
$\alpha +\beta +\gamma +\partial =\dfrac{-b}{a}$
Since the coefficients are rational we have $\dfrac{-b}{a}$ is rational.
Hence the sum of the roots must also be rational.
Now let us check the options
If we choose option d then the roots of the expression are $\sqrt{2}+\sqrt{3},\sqrt{2}-\sqrt{3},-\sqrt{2}+\sqrt{3},-\sqrt{2}-\sqrt{3}$
In this case we will get the sum of the roots as 0 and hence rational.
So, the correct answer is “Option d”.
Note: Now from the relation between the coefficient and the roots we also have $\alpha \beta +\beta \gamma +\gamma \partial +\partial \alpha +\beta \partial +\alpha \gamma =\dfrac{c}{a}$, $\alpha \beta \gamma +\beta \gamma \partial +\gamma \partial \alpha =\dfrac{-d}{a}$ and the product of the roots which is $\alpha \beta \gamma \partial $ is $\dfrac{e}{a}$. We have the result generalized for polynomials of degree n.
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