Question

The total number of integral values of $a$ so that ${{x}^{2}}-\left( a+1 \right)x+a-1=0$ has integral roots is equal to
A. $1$
B. $2$
C. $4$
D. None of these

Answer 1

Hint: In this problem we have to find the number of values of $a$ so that the equation ${{x}^{2}}-\left( a+1 \right)x+a-1=0$ has integral root. We know that the roots of a quadratic equation $p{{x}^{2}}+qx+r=0$ are given by $x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}$. From the above formula we can say that ‘If a quadratic equation has integral roots then the value of ${{q}^{2}}-4pr$ should be a perfect square. So, we will compare the given equation ${{x}^{2}}-\left( a+1 \right)x+a-1=0$ with $p{{x}^{2}}+qx+r=0$. Now we will write the value ${{q}^{2}}-4pr$ in terms of $a$. From the given condition, we will check for the values of $a$ for which the equation ${{q}^{2}}-4pr$ gives a perfect square.

Complete step-by-step answer:
Given Equation is ${{x}^{2}}-\left( a+1 \right)x+a-1=0$
Comparing the given equation with $p{{x}^{2}}+qx+r=0$, then we will get the values of $p,q,r$ as
$p=1$, $q=-\left( a+1 \right)$, $r=a-1$
Now the value of ${{q}^{2}}-4pr$ is given by
${{q}^{2}}-4pr={{\left[ -\left( a+1 \right) \right]}^{2}}-4\left( 1 \right)\left( a-1 \right)$
We know that ${{\left( -a \right)}^{2}}={{a}^{2}}$ so ${{\left[ -\left( a+1 \right) \right]}^{2}}={{\left( a+1 \right)}^{2}}$, then the above equation modified as
${{q}^{2}}-4pr={{\left( a+1 \right)}^{2}}-4\left( a-1 \right)$
Expanding the term ${{\left( a+1 \right)}^{2}}$ by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ where $a=a$, $b=1$ then we will get
$\begin{align}
  & {{q}^{2}}-4pr={{a}^{2}}+{{1}^{2}}+2\left( a \right)\left( 1 \right)-4a+4 \\
 & \Rightarrow {{q}^{2}}-4pr={{a}^{2}}+1+2a-4a+4 \\
 & \Rightarrow {{q}^{2}}-4pr={{a}^{2}}-2a+5 \\
\end{align}$
Here we have the value of ${{q}^{2}}-4pr$ as ${{a}^{2}}-2a+5$. If the given equation has an integral root, then the value of ${{q}^{2}}-4pr$ should be a perfect square. The value of ${{q}^{2}}-4pr$ is a quadratic equation in terms of $a$. So, let us substituting the value of $a=0,\pm 1,\pm 2,...$ up to where we will get the value of equation ${{a}^{2}}-2a+5$ as perfect square.
$\therefore $ Substituting the value of $a=0$, then the value of ${{a}^{2}}-2a+5$ will be
$\begin{align}
  & {{a}^{2}}-2a+5={{0}^{2}}-2\left( 0 \right)+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=5 \\
\end{align}$
Here the value $5$ which is not a perfect square.
For $a=1$ the value of ${{a}^{2}}-2a+5$ will be
$\begin{align}
  & {{a}^{2}}-2a+5={{1}^{2}}-2\left( 1 \right)+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=1-2+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=4 \\
\end{align}$
Here the value $4$ which is a perfect square.
Hence $a$ may have at least one integral value that stratifies the given condition.
For $a=-1$ the value of ${{a}^{2}}-2a+5$ will be
$\begin{align}
  & {{a}^{2}}-2a+5={{\left( -1 \right)}^{2}}-2\left( -1 \right)+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=1+2+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=8 \\
\end{align}$
Here the value $8$ which is not a perfect square.
For $a=2$ the value of ${{a}^{2}}-2a+5$ will be
$\begin{align}
  & {{a}^{2}}-2a+5={{2}^{2}}-2\left( 2 \right)+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=4-4+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=5 \\
\end{align}$
Here the value $5$ which is not a perfect square.
For $a=-2$ the value of ${{a}^{2}}-2a+5$ will be
$\begin{align}
  & {{a}^{2}}-2a+5={{\left( -2 \right)}^{2}}-2\left( -2 \right)+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=4+4+5 \\
 & \Rightarrow {{a}^{2}}-2a+5=13 \\
\end{align}$
Here the value $13$ which is not a perfect square.
From all the above values, we only have value of $a=1$ where the equation ${{x}^{2}}-\left( a+1 \right)x+a-1=0$has integral root. Hence the number of integral values of $a$ is One.
$\therefore $ Option – A is the correct one.

So, the correct answer is “Option A”.

Note: Students may get confused with the terminology used in the question and they try to find the value of $a$ that satisfies ${{q}^{2}}-4qr=0$. If you get the values of $a$ that satisfies ${{q}^{2}}-4qr=0$ is not the correct choice because for the values ${{q}^{2}}-4qr>0$ the quadratic equation has integral roots. So, we need to take the condition that the value of ${{q}^{2}}-4pr$ should be a perfect square.