Question

Find the number of pairs $\left( a,b \right)$ of real numbers such that whenever $\alpha $ is a root of ${{x}^{2}}+ax+b=0,{{\alpha }^{2}}-2$ is also a root of the equation.
A. 4
B. 3
C. 5
D. 6

Answer 1

Hint: For solving this question you should know about the roots of a quadratic equation and the sum of the roots and product of the roots also. In the problem first we will find the roots of a quadratic equation which is given in the question and then we will find the sum of the given roots and product of their roots and then will find all the real numbers.

Complete step by step answer:
According to our question it is asked to us to find the number of all pairs of real numbers for a given quadratic equation and a root is already given for that. The quadratic equation given is: ${{x}^{2}}+ax+b=0$ and if we consider the first root as $\alpha $ and second root as $\beta $.
As we know there are two possible situation, for the occurrence of roots
Case I : when we have $\alpha =\beta $
So, if we consider this case, we get the following situation
As we have been given that ${{\alpha }^{2}}-2$ and $\alpha $ are the roots, so we can say
${{\alpha }^{2}}-2=\alpha $
Now we will use middle term splitting to split these terms, so we get
$\begin{align}
  & {{\alpha }^{2}}-\alpha -2=0 \\
 & \Rightarrow {{\alpha }^{2}}-2\alpha +\alpha -2=0 \\
 & \Rightarrow \alpha \left( \alpha -2 \right)+1\left( \alpha -2 \right)=0 \\
 & \Rightarrow \left( \alpha -2 \right)\left( \alpha +1 \right)=0 \\
 & \Rightarrow \alpha =2\text{ and }\alpha =-1 \\
\end{align}$
And by the properties of sum and product of zeroes of quadratic equation, we get
$a=-2\alpha \text{ and b=}{{\alpha }^{2}}$
Therefore, we get values of (a, b) as (2, 1) and (-4, 4).
Case II : when we have $\alpha \ne \beta $, so we have four possibilities
(i) \[\alpha ={{\alpha }^{2}}-2\] and \[\beta ={{\beta }^{2}}-2\]
By solving both the above equation individually, we get two pairs of values of $\left( \alpha ,\beta \right)$, and they are $\left( \alpha ,\beta \right)=\left( 2,-1 \right)\text{ and }\left( -1,2 \right)$
And by the properties of sum and product of zeroes of quadratic equation, we get
$a=-\left( \alpha +\beta \right)\text{ and b=}\alpha \beta $
Therefore, we get values of (a, b) as (-1, -2).
(ii) \[\alpha ={{\beta }^{2}}-2\] and \[\beta ={{\alpha }^{2}}-2\]
Now, we will subtract \[\alpha ={{\beta }^{2}}-2\] from \[\beta ={{\alpha }^{2}}-2\], we get
\[\begin{align}
  & \beta -\alpha ={{\alpha }^{2}}-{{\beta }^{2}} \\
 & \Rightarrow \beta -\alpha =\left( \alpha -\beta \right)\left( \alpha +\beta \right) \\
\end{align}\]
Therefore, we get \[\alpha +\beta =-1\].
We also have
\[\begin{align}
  & \beta +\alpha ={{\alpha }^{2}}+{{\beta }^{2}}-4 \\
 & \Rightarrow \beta +\alpha ={{\left( \beta +\alpha \right)}^{2}}-2\beta \alpha -4 \\
 & \Rightarrow -1={{\left( -1 \right)}^{2}}-2\beta \alpha -4 \\
 & \Rightarrow -1=1-2\beta \alpha -4 \\
 & \Rightarrow \alpha \beta =-1 \\
\end{align}\]
And by the properties of sum and product of zeroes of quadratic equation, we get
$a=-\left( \alpha +\beta \right)\text{ and b=}\alpha \beta $
Therefore, we get values of (a, b) as (1, -1).
(iii) \[\alpha ={{\alpha }^{2}}-2={{\beta }^{2}}-2\] and \[\alpha \ne \beta \]
Therefore, we can say \[\alpha =-\beta \]
By solving both the above equation as \[\alpha ={{\alpha }^{2}}-2\] and \[\alpha =-\beta \], we get two pairs of values of $\left( \alpha ,\beta \right)$, and they are $\left( \alpha ,\beta \right)=\left( 2,-2 \right)\text{ and }\left( -1,1 \right)$
And by the properties of sum and product of zeroes of quadratic equation, we get
$a=-\left( \alpha +\beta \right)\text{ and b=}\alpha \beta $
Therefore, we get values of (a, b) as (0, -4) and (0,-1).
(iv) \[\beta ={{\alpha }^{2}}-2={{\beta }^{2}}-2\] and \[\alpha \ne \beta \]
Therefore, we can say \[\alpha =-\beta \]
By solving both the above equation as \[\beta ={{\beta }^{2}}-2\] and \[\alpha =-\beta \], we get two pairs of values of $\left( \alpha ,\beta \right)$, and they are $\left( \alpha ,\beta \right)=\left( 2,-2 \right)\text{ and }\left( -1,1 \right)$
And by the properties of sum and product of zeroes of quadratic equation, we get
$a=-\left( \alpha +\beta \right)\text{ and b=}\alpha \beta $
Therefore, we get values of (a, b) as (0, -4) and (0,-1).

Hence we get 6 pairs of real numbers (a, b) as following:
(-4, 4), (2, 1) , (-1, -2) , (1, -1) , (0, -4) , (0, -1)

So, the correct answer is “Option D”.

Note: While solving these types of questions you have to be careful if the roots are given already because here all pairs will exist if both are real numbers like as it in many cases it can be possible that this condition will not satisfy. So, first find some solutions for that and then make it right at the final process.