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

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 $f(x) = 0$.