Question

If we have $\alpha ,\beta ,\gamma ,\delta$ are the roots of equation ${{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d+e=0$, find the value of $\sum{{{\alpha }^{2}}\beta }$.

Answer 1

Here we are given an equation of degree four, thus having our roots. We will find the sum and product of roots in terms of coefficients of the equation to find desired results. For equation of degree four, ${{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d+e=0$, sum of roots is given as –
$\alpha +\beta +\gamma +\delta =-\dfrac{b}{a}$ .
Product of roots is given as –
$\alpha \beta \gamma \delta =\dfrac{e}{a}$.
Also, $\alpha \beta +\beta \gamma +\gamma \delta +\alpha \gamma +\alpha \delta +\beta \delta =\dfrac{c}{a}$ and
$\alpha \beta \gamma +\alpha \gamma \delta +\alpha \beta \delta +\gamma \beta \delta =-\dfrac{d}{a}$.
We will use these formulas for finding $\sum{{{\alpha }^{2}}\beta }$.

Complete step-by-step solution
Before applying direct formulas and jumping to answer, let us first understand the basic formulas for ${{n}^{th}}$ polynomial.
For a polynomial of degree $n$, let roots of equation are $\alpha ,{{\alpha }_{1}},{{\alpha }_{2}},...,{{\alpha }_{n}}$.
Equation in general form is given by –
$f\left( x \right)={{a}_{0}}{{x}^{n}}+{{a}_{1}}{{x}^{n-1}}+{{a}_{2}}{{x}^{n-2}}+...+{{a}_{n-1}}x+{{a}_{n}}=0$
Then,
Sum of roots, $\alpha +{{\alpha }_{1}}+{{\alpha }_{2}}+...+{{\alpha }_{n}}=\dfrac{-coefficient~~of~~{{x}^{n-1}}}{coefficient~~of~~{{x}^{n}}}$
Also, ${{\alpha }_{1}}{{\alpha }_{2}}+{{\alpha }_{1}}{{\alpha }_{3}}...={{\left( -1 \right)}^{2}}\dfrac{coefficient~~of~~{{x}^{n-2}}}{coefficient~~of~~{{x}^{n}}}$
Similarly, other formulas are:-
${{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}+{{\alpha }_{2}}{{\alpha }_{3}}{{\alpha }_{4}}...={{\left( -1 \right)}^{3}}\dfrac{coefficient~~of~~{{x}^{n-3}}}{coefficient~~of~~{{x}^{n}}}$
${{\alpha }_{1}}{{\alpha }_{2}}{{\alpha }_{3}}{{\alpha }_{4}}...{{\alpha }_{n}}={{\left( -1 \right)}^{n}}\dfrac{constant~~term}{coefficient~~of~~{{x}^{n}}}$
Comparing general formulas by the general equation of degree four, ${{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d+e=0$ having roots $\alpha ,\beta ,\gamma ,\delta$ as roots:

& \alpha +\beta +\gamma +\delta =-\dfrac{b}{a} \\\ & \alpha \beta +\beta \gamma +\gamma \delta +\alpha \gamma +\alpha \delta +\beta \delta =\dfrac{c}{a} \\\ & \alpha \beta \gamma +\alpha \gamma \delta +\alpha \beta \delta +\gamma \beta \delta =-\dfrac{d}{a} \\\ & \alpha \beta \gamma \delta =\dfrac{e}{a} \\\ \end{aligned}$$ We are given the equation, $${{x}^{4}}+a{{x}^{3}}+b{{x}^{2}}+cx+d+e=0$$. Comparing with above formulas we get – $$\begin{aligned} & \alpha +\beta +\gamma +\delta =-a~~~~~~~~~~~~~~~~~~~~~~~~~~~~...\left( 1 \right) \\\ & \alpha \beta +\beta \gamma +\gamma \delta +\alpha \gamma +\alpha \delta +\beta \delta =b~~~~~...\left( 2 \right) \\\ & \alpha \beta \gamma +\alpha \gamma \delta +\alpha \beta \delta +\gamma \beta \delta =-c~~~~~~~~~~~~...\left( 3 \right) \\\ & \alpha \beta \gamma \delta =e \\\ \end{aligned}$$ On multiplying $\left( 1 \right)$ and $\left( 2 \right)$, we get – $$\begin{aligned} & \left( \alpha +\beta +\gamma +\delta \right)\left( \alpha \beta +\beta \gamma +\gamma \delta +\alpha \gamma +\alpha \delta +\beta \delta \right)=-ab \\\ & \Rightarrow {{\alpha }^{2}}\beta +\alpha {{\beta }^{2}}+\alpha \beta \gamma +\alpha \beta \delta +\alpha \beta \gamma +{{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}+\beta \gamma \delta +\alpha \gamma \delta +\beta \gamma \delta +{{\gamma }^{2}}\delta +\gamma {{\delta }^{2}}+ \\\ & {{\alpha }^{2}}\gamma +\alpha \beta \gamma +\alpha {{\gamma }^{2}}+\alpha \gamma \delta +{{\alpha }^{2}}\delta +\alpha \beta \delta +\alpha \gamma \delta +\alpha {{\delta }^{2}}+\alpha \beta \delta +{{\beta }^{2}}\delta +\gamma \beta \delta +\beta {{\delta }^{2}}=-ab \\\ \end{aligned}$$ Rearranging the terms, we get – $$\begin{aligned} & {{\alpha }^{2}}\beta +\alpha {{\beta }^{2}}+{{\beta }^{2}}\gamma +\beta {{\gamma }^{2}}+{{\gamma }^{2}}\delta +\gamma {{\delta }^{2}}+{{\alpha }^{2}}\gamma +\alpha {{\gamma }^{2}}+{{\alpha }^{2}}\delta +\alpha {{\delta }^{2}}+{{\beta }^{2}}\delta +\beta {{\delta }^{2}}+ \\\ & 3\left( \alpha \beta \gamma +\beta \gamma \delta +\alpha \beta \delta +\beta \gamma \delta \right)=-ab \\\ \end{aligned}$$ We can write the first twelve terms by $\sum{{{\alpha }^{2}}\beta }$. Hence we get – $\sum{{{\alpha }^{2}}\beta }+3\left( \alpha \beta \gamma +\beta \gamma \delta +\alpha \beta \delta +\beta \gamma \delta \right)=-ab$ From equation (3), we can clearly see that we can directly put the value of $-c$ in above equation. We get – $$\begin{aligned} & \sum{{{\alpha }^{2}}\beta }-3c=-ab \\\ & \Rightarrow \sum{{{\alpha }^{2}}\beta }=3c-ab \\\ \end{aligned}$$ **Hence, we have found our required answer, which is, $$\sum{{{\alpha }^{2}}\beta }=3c-ab$$.** **Note:** Students should take care of signs the most. Mistakes can be made while taking positive or negative coefficients. Easy way to remember the product of roots is that we take the positive value of the coefficient of the constant term in even polynomials and the negative value of the coefficient of the constant term in odd polynomials. As there are a lot of terms in the equation, students should do it carefully and do not skip any term. Always remember, we can check it by looking at the symmetry of terms.