Question

If $\alpha \ and\ \beta $ are the zeros of the quadratic polynomial \[f\left( x \right) = a{{x}^{2}}+bx+c\], then evaluate: $\alpha -\beta $.

Answer 1

Hint: We will be using the concept of quadratic equations to solve the problem. We will be using the concept of sum of zeroes and product of zeroes also we will be using algebraic identities like,
$\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}$

Complete Step-by-Step solution:
Now, we have been given $\alpha \ and\ \beta $are the zeros of the polynomial\[f\left( x \right)=a{{x}^{2}}+bx+c\]. We have to evaluate the value of $\alpha -\beta $.
Now, we know that in a quadratic polynomial $a{{x}^{2}}+bx+c$.
$\begin{align}
  & \text{sum of zeros }=\dfrac{-b}{a} \\
 & \text{product of zeros }=\dfrac{c}{a} \\
\end{align}$
Therefore, we have the value of,
$\begin{align}
  & \alpha +\beta =\dfrac{-b}{a} \\
 & \alpha \beta =\dfrac{c}{a} \\
\end{align}$
Now, we have to find the value of $\alpha -\beta $ for this we will write ${{\left( \alpha -\beta \right)}^{2}}$ in terms of $\alpha +\beta \ and\ \alpha \beta $ as,
${{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}-2\alpha \beta $
Now, we will add and subtract $2\alpha \beta $ from RHS.
$\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta -4\alpha \beta \\
 & {{\left( \alpha -\beta \right)}^{2}}={{\left( \alpha +\beta \right)}^{2}}-4\alpha \beta \\
\end{align}$
Now, we will substitute the value of $\alpha +\beta \ and\ \alpha \beta $ from (1) and (2).
$\begin{align}
  & {{\left( \alpha -\beta \right)}^{2}}={{\left( \dfrac{-b}{a} \right)}^{2}}-\dfrac{4c}{a} \\
 & =\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{4c}{a} \\
 & =\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}} \\
 & {{\left( \alpha -\beta \right)}^{2}}=\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}} \\
 & \alpha -\beta =\pm \sqrt{\dfrac{{{b}^{2}}-4ac}{{{a}^{2}}}} \\
 & =\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{a} \\
\end{align}$
Therefore, the value of $\alpha -\beta =\pm \dfrac{\sqrt{{{b}^{2}}-4ac}}{a}$.

Note: To solve these types of questions it is important to remember the relation between coefficient of quadratic equation and roots of the quadratic equation. Also identities like,
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\ and\ {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ are also important.