Question

Find the value of $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)$ given that $\alpha \text{ and }\beta $ are roots of the given quadratic equation: ${{x}^{2}}-3x+1$ .
A). 3
B). 4
C). 6
D). 8

Answer 1

Hint: As we have given the equation we can use the formula of finding the sum of roots and product of roots and from using these two equations we can calculate the value that is asked to find.

Complete step-by-step solution
Now we are going to use the formula that will work for any polynomial equation for finding the product of roots of the given equation,
The formula is:
If the given equation is $a{{x}^{2}}+bx+c$ then ,
Product of root = $\dfrac{c}{a}$
Sum of root = $\dfrac{-b}{a}$
So, now using this formula,
$\begin{align}
  & \alpha \beta =\dfrac{1}{1} \\
 &\Rightarrow \alpha +\beta =\dfrac{-\left( -3 \right)}{1} \\
\end{align}$
Now we will use this formula: ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
Therefore,
$\begin{align}
  & {{\left( \alpha +\beta \right)}^{2}}={{\alpha }^{2}}+{{\beta }^{2}}+2\alpha \beta \\
 &\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left( \alpha +\beta \right)}^{2}}-2\alpha \beta \\
\end{align}$
Now we will substitute all the values that we have calculated above,
After substituting the values we get,
$\begin{align}
  &\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}={{\left( 3 \right)}^{2}}-2\times 1 \\
 &\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=9-2 \\
 &\Rightarrow {{\alpha }^{2}}+{{\beta }^{2}}=7 \\
\end{align}$
Now the value of $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)$ will be
$= 7 + 1$
$= 8$
Hence, the correct answer is 8. So, the correct option is (d).

Note: We should always try to use the given formula to solve the question as it saves time and reduces the probability of making mistakes. Another method which one can use is to find all the roots of the given quadratic equation and then we can put these values in $\left( {{\alpha }^{2}}+{{\beta }^{2}}+\alpha \beta \right)$and then we will get the desired result that we want.