An equation of the form 

\({a_0}{x^n} + {a_1}{x^{n - 1}}^{} + {a_2}{x^{n - 2}} + ......... + {a_n} = 0\) 

is called an \(nth\) degree polynomial equation. 

There will be \(n\) values of \(x\), real or imaginary, satisfying the given equation. These values are called the roots of the equation. If \({x_1},\,{x_2}.........{x_n}\) be the roots of equation, then the equation can be written as:

\({a_0}(x - {x_1})(x - {x_2})......(x - {x_n}) = 0\)

for we can see that \({x_1},{x_2}, \ldots  \ldots  \ldots  \ldots  \ldots ..{x_n}\) satisfy the above equation.

Now we have the following results:

Sum of the roots taken one at a time =\( - \left( {\frac{{{a_1}}}{{{a_0}}}} \right)\)

Sum of the roots taken two at a time =\( + \left( {\frac{{{a_2}}}{{{a_0}}}} \right)\)

Sum of the roots taken \(n\) at a time = \({\left( {-1} \right)^n}\) \(\left( {\frac{{{a_n}}}{{{a_0}}}} \right)\)

Example 15: Let \(\alpha \) and \(\,\beta \) be the roots of the equation \({x^3} + {\rm{ }}a{x^2} + {\rm{ }}bx{\rm{ }} + {\rm{ }}c{\rm{ }} = {\rm{ }}0\) satisfying the relation \(\alpha \beta  + 1=0\). Prove that \({c^2} + {\rm{ }}ac{\rm{ }} + {\rm{ }}b{\rm{ }} + {\rm{ }}1 = {\rm{ }}0\).

Solution:  If \(\alpha ,\,\;\beta \) and \(\gamma \)be the roots of the given equation, then we have 

\(\alpha  + \beta  + \gamma  = {\rm{ }}-a\) …………………….. (1)

\(\alpha \beta  + \alpha \gamma  + \beta \gamma  = b\) …………………….. (2)

and \(\alpha \beta \gamma  =  - c\) …………………….. (3)

Also, we have \(\alpha \beta  + 1 = 0\)

Putting \(\alpha \beta  =  - 1\)in the equation (3) we have, \(\gamma  = c\)  

putting the value of \(\gamma  = c\)

Putting the value of \(\gamma \) in equation (1), we have 

\(\alpha  + \beta  =  - a - c\)

Now, putting the above values in equation (2), we have: \(\alpha \beta  + \gamma (\alpha  + \beta ) = b\)

i.e.  \(-1 - c(a + c) = b\)

i.e. \({c^2} + ac + b + {\rm{ }}1{\rm{ }} = {\rm{ }}0\) which is the desired result.

Example 16: For the quadratic equation 

\(\left( {m{\rm{ }}-{\rm{ }}2} \right){x^2} + {\rm{ }}\left( {2m{\rm{ }}-{\rm{ }}8} \right)x{\rm{ }} + {\rm{ }}\left( {3m{\rm{ }}-{\rm{ }}8} \right){\rm{ }} = {\rm{ }}0\), find the values of m for which 

(a) The roots are real.

(b) The roots are of opposite sign.

Solution: (a) For roots to be real when \(D \ge 0\)

i.e. \({\left( {2m-{\rm{ }}8} \right)^2} \ge 4\left( {m-{\rm{ }}2} \right)\left( {3m-{\rm{ }}8} \right)\)

i.e. \({m^2}-{\rm{ }}8m + {\rm{ }}16 \ge 3{m^2}-{\rm{ }}14m + {\rm{ }}16\)

i.e. \(2{m^2}-{\rm{ }}6m\; \le 0 \Rightarrow \;m\left( {m-{\rm{ }}3} \right) \le 0\)

\( \Rightarrow \) \(0 \le m \le 3\).

(b) For roots to be of opposite sign, product of roots  must be negative

\(\frac{{3m - 8}}{{m - 2}} < 0\) \( \Rightarrow \) m belongs to the interval\(\left( {2,\,\,\,\frac{8}{3}} \right)\)

Example 17: For the quadratic equation, 

\({x^2}-{\rm{ }}\left( {m-{\rm{ }}3} \right)x + m = {\rm{ }}0\), find the values of \(m\) for which the roots are real and distinct?

Solution: If  \({b^2}-{\rm{ }}4ac = D\), for real and distinct roots \(D > 0\)    

i.e. \({\left( {m-{\rm{ }}3} \right)^2}-{\rm{ }}4m > {\rm{ }}0 \Rightarrow {m^2}-{\rm{ }}10m + {\rm{ }}9{\rm{ }} > {\rm{ }}0\)

i.e. \(\left( {m-{\rm{ }}1} \right)\left( {m-{\rm{ }}9} \right){\rm{ }} > {\rm{ }}0 \Rightarrow 1{\rm{ }} < m < {\rm{ }}9\)