Question

If \[{p^3} - q\left( {3p - 1} \right) + {q^2} = 0\], find the relation between the roots of the equation \[{x^2} + px + q = 0\].

Answer 1

Hint: We will assume the roots of the equation \[{x^2} + px + q = 0\] to be \[\alpha \] and \[\beta \]. As we know, standard form of an quadratic equation is \[a{x^2} + bx + c\], where \[a\], \[b\] and \[c\] are the coefficients and the quadratic equation in term of roots is given by: \[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\]. Using this we will find the value of sum of the roots and product of the roots in terms of \[p\] and \[q\]. Then we will substitute this in the given equation \[{p^3} - q\left( {3p - 1} \right) + {q^2} = 0\] to find the relation between the roots of the equation \[{x^2} + px + q = 0\].

Complete step-by-step answer:
Let \[\alpha \] and \[\beta \] are the roots of the equation \[{x^2} + px + q = 0\].
As we know, standard form of an quadratic equation is \[a{x^2} + bx + c\], where \[a\], \[b\] and \[c\] are the coefficients and the quadratic equation in term of roots is given by: \[{x^2} - \left( {{\text{sum of the roots}}} \right)x + \left( {{\text{product of the roots}}} \right) = 0\].
Using this, we get
\[ \Rightarrow \alpha + \beta = - p\] and \[\alpha \beta = q\]
Given, \[{p^3} - q\left( {3p - 1} \right) + {q^2} = 0\].
Putting the values of \[p\] and \[q\] in the above equation, we get
\[ \Rightarrow {\left( { - \left( {\alpha + \beta } \right)} \right)^3} - \alpha \beta \left( { - 3\left( {\alpha + \beta } \right) - 1} \right) + {\left( {\alpha \beta } \right)^2} = 0\]
On simplifying, we get
\[ \Rightarrow - \left( {{\alpha ^3} + 3{\alpha ^2}\beta + 3\alpha {\beta ^2} + {\beta ^3}} \right) - \alpha \beta \left( { - 3\alpha - 3\beta - 1} \right) + {\alpha ^2}{\beta ^2} = 0\]
On further simplification, we get
\[ \Rightarrow - {\alpha ^3} - 3{\alpha ^2}\beta - 3\alpha {\beta ^2} - {\beta ^3} + 3{\alpha ^2}\beta + 3\alpha {\beta ^2} + \alpha \beta + {\alpha ^2}{\beta ^2} = 0\]
\[ \Rightarrow - {\alpha ^3} - {\beta ^3} + \alpha \beta + {\alpha ^2}{\beta ^2} = 0\]
On rewriting, we get
\[ \Rightarrow \alpha \beta + {\alpha ^2}{\beta ^2} = {\alpha ^3} + {\beta ^3}\]
Taking common from the left-hand side of the above equation, we get
\[ \Rightarrow \alpha \beta \left( {\alpha \beta + 1} \right) = {\alpha ^3} + {\beta ^3}\]
Therefore, the relation between the roots of the equation \[{x^2} + px + q = 0\], given \[{p^3} - q\left( {3p - 1} \right) + {q^2} = 0\] is \[\alpha \beta \left( {\alpha \beta + 1} \right) = {\alpha ^3} + {\beta ^3}\].

Note: \[{x^2} + px + q = 0\] is a quadratic equation. A quadratic equation function may have one, two, or zero roots. Roots are also called the x-intercept or zeroes. Also, the y-coordinate of any points lying on the x-axis is zero. So, to find the roots of a quadratic function, we set \[f(x) = 0\].