Question

If α and β are the zeroes of a Quadratic polynomial${{x}^{2}}+3x+6$. Find the values of
$\begin{align}
  & 1)\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha } \\
 & 2){{\alpha }^{2}}+{{\beta }^{2}} \\
\end{align}$

Answer 1

Hint: we know if α and β are the roots of the equation $a{{x}^{2}}+bx+c=0$ then the sum of roots α + β is given by $\dfrac{-b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$. Now once we have α+β and αβ we can write the required terms in the form of α+β and αβ with the formula ${{a}^{2}}+{{b}^{2}}+2ab={{(a+b)}^{2}}$ and hence find the value.

Complete step-by-step answer:
Now the given equation is ${{x}^{2}}+3x+6$
Now we know if α and β are the roots of the equation $a{{x}^{2}}+bx+c=0$ then the sum of roots α + β is given by $\dfrac{-b}{a}$ and the product of the roots is given by $\dfrac{c}{a}$
Hence comparing it with the equation ${{x}^{2}}+3x+6$ we get
$\begin{align}
  & \alpha +\beta =\dfrac{-3}{1}=-3.............(1) \\
 & \alpha \beta =\dfrac{6}{1}=6.....................(2) \\
\end{align}$
Now let us find $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }$
We will try to write this in the form of $\alpha +\beta $ and $\alpha \beta $
$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }$ {cross multiplying the fractions}
Now ${{a}^{2}}+{{b}^{2}}+2ab={{(a+b)}^{2}}$ using this in above equation we get
$\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }$
Now substituting the values from equation (1) and equation (2) in the above equation we get
\[\begin{align}
  & \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{{{(-3)}^{2}}-2(6)}{6} \\
 & \Rightarrow \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{9-12}{6} \\
 & \Rightarrow \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=\dfrac{-3}{2} \\
 & \Rightarrow \dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=-\dfrac{1}{2} \\
\end{align}\]
Hence we get $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=-\dfrac{1}{2}................(3)$
Now Consider ${{\alpha }^{2}}+{{\beta }^{2}}$
We know that ${{a}^{2}}+{{b}^{2}}+2ab={{(a+b)}^{2}}$
This means\[{{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta ={{(\alpha +\beta )}^{2}}\]
Hence \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta \]
Now substituting the values from equation (1) and equation (2) we get.
$\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}={{(-3)}^{2}}-2(6) \\
 & {{\alpha }^{2}}+{{\beta }^{2}}=9-12 \\
 & {{\alpha }^{2}}+{{\beta }^{2}}=-3 \\
\end{align}$
Hence we get the value of ${{\alpha }^{2}}+{{\beta }^{2}}=-3.................(4)$
Hence from equation (3) and equation (4) we have the required values that is
${{\alpha }^{2}}+{{\beta }^{2}}=-3.$ and $\dfrac{\alpha }{\beta }+\dfrac{\beta }{\alpha }=-\dfrac{1}{2}.$

Note: Now instead of finding the sum of roots and products of roots and using their substitution we can also find the solution directly
Now we know for any quadratic equation $a{{x}^{2}}+bx+c=0$ we have the roots are equal to $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ hence, once we have both the roots we can substitute the values of the roots in the given expression and solve.