Question

If $ \alpha ,\beta ,\gamma ,\delta $ are the roots of the equation $ {x^4} + q{x^2} + rx + s = 0 $ find the equation whose roots are $ \beta + \gamma + \delta + {(\beta \gamma \delta )^{ - 1}} $ , and c.

Answer 1

Hint: we will use the concept of transformation of equations for which we have to know the relation between the roots and the required roots and then apply the transformation. Firstly, we will find the relation between roots and the coefficients of the equations and then use those relations to find the required equation.

Complete step-by-step answer:
Now, from the question
Given equation, $ {x^4} + q{x^2} + rx + s = 0 $ which has the roots $ \alpha ,\beta ,\gamma ,\delta $
Therefore, $ \alpha + \beta + \gamma + \delta = \dfrac{{ - b}}{a} = 0 $
 $ \alpha \beta + \beta \gamma + \gamma \delta + \alpha \gamma + \alpha \delta + \beta \delta = \dfrac{c}{a} = q $
 $ \alpha \beta \gamma + \alpha \gamma \delta + \beta \gamma \delta + \alpha \beta \delta = \dfrac{{ - d}}{a} = - r $
 $ \alpha \beta \gamma \delta = \dfrac{e}{a} = s $
Now, $ \beta + \gamma + \delta + \dfrac{1}{{\beta \gamma \delta }} = (\alpha + \beta + \gamma + \delta ) + \dfrac{\alpha }{{\alpha \beta \gamma \delta }} - \alpha $ [adding and subtracting $ \alpha $ everywhere]
Substituting values of the above, we get
 $ (\dfrac{{1 - s}}{s})\alpha $
Therefore, roots of the required equation are
 $ (\dfrac{{1 - s}}{s})\alpha ,(\dfrac{{1 - s}}{s})\beta ,(\dfrac{{1 - s}}{s})\gamma ,(\dfrac{{1 - s}}{s})\delta ,c $
Or roots of the required equation are
 $ \lambda \alpha ,\lambda \beta ,\lambda \gamma ,\lambda \delta ,c $ where $ \lambda = \dfrac{{1 - s}}{s} $
For the required equation,
 $ {S_1} = \lambda \alpha + \lambda \beta + \lambda \gamma + \lambda \delta + c = \lambda (\alpha + \beta + \gamma + \delta ) + c = c $ [since, $ \alpha + \beta + \gamma + \delta = \dfrac{{ - b}}{a} = 0 $ ]
 $ {S_2} = {\lambda ^2}\sum\limits_{}^{} {\alpha \beta + \lambda (\sum\limits_{}^{} \alpha )c = {\lambda ^2}q} $ [ $ \alpha \beta + \beta \gamma + \gamma \delta + \alpha \gamma + \alpha \delta + \beta \delta = \dfrac{c}{a} = q $ ]
 $ {S_3} = {\lambda ^3}\sum\limits_{}^{} {\alpha \beta \gamma } + {\lambda ^2}(\sum\limits_{}^{} {\alpha \beta } )c = - {\lambda ^3}r + {\lambda ^2}rc $ [ $ \alpha \beta \gamma + \alpha \gamma \delta + \beta \gamma \delta + \alpha \beta \delta = \dfrac{{ - d}}{a} = - r $ ]
 $ {S_4} = {\lambda ^4}\alpha \beta \gamma \delta + {\lambda ^3}(\sum\limits_{}^{} {\alpha \beta \gamma )c = {\lambda ^4}s = {\lambda ^3}rc} $ [ $ \alpha \beta \gamma \delta = \dfrac{e}{a} = s $ ]
 $ {S_5} = {\lambda ^4}\alpha \beta \gamma \delta c = {\lambda ^4}sc $ [ $ \alpha \beta \gamma \delta = \dfrac{e}{a} = s $ ]
Required equation is given by,
 $ {x^5} - {S_1}{x^4} + {S_2}{x^3} - {S_3}{x^2} + {S_4}x - {S_5} = 0 $
Or, $ {x^5} - c{x^4} + {\lambda ^2}q{x^3} - ( - {\lambda ^3}r + {\lambda ^2}rc){x^2} + ({\lambda ^4}s - {\lambda ^3}rc)x - {\lambda ^4}sc = 0 $
Which is the required equation

Note: Linear Polynomial
The linear polynomial is an expression, in which the degree of the polynomial is 1. The linear polynomial should be within the sort of ax+b. Here, “x” may be a variable, “a” and “b” are constant.
The polynomial P(x) is ax+b, then the zero of a polynomial is $ \dfrac{{ - b}}{a} $
Quadratic Polynomial
The Quadratic polynomial is defined as a polynomial with the highest degree of 2. The quadratic polynomial should be in the form of $ a{x^2} + bx + c = 0 $ . In this case, a ≠ 0. Let say α and β are the 2 zeros of a polynomial, then
The sum of zeros, α + β is $ \dfrac{{ - b}}{a} $
The product of zeros, αβ is $ \dfrac{c}{a} $
Cubic Polynomial
The cubic polynomial is a polynomial with the highest degree of 3. The cubic polynomial should be within the sort of $ a{x^2} + bx + c + d = 0 $ , where a ≠ 0. Let say α, β, and γ are the three zeros of a polynomial, then
The sum of zeros, α + β + γ is $ \dfrac{{ - b}}{a} $
The sum of the merchandise of zeros, αβ+ βγ + αγ is $ \dfrac{c}{a} $
The product of zeros, αβγ is $ \dfrac{{ - d}}{a} $