Question

If \[({a^2} - 1){x^2} + (a - 1)x + {a^2} - 4a + 3 = 0\] be an identity in x then the value of a is/are?
A) -1
B) 1
C) 3
D) -1,1,3

Answer 1

Hint: The given equation is identical in x, therefore the coefficients must be zero. Use this property to get 3 equations and then solve them to find the common value of a in all 3 of them.

Complete Step by Step Solution:
 If we compare the given equation with the general quadratic equation which is
\[A{x^2} + Bx + C = 0\]
For this equation to be identical in x, A B and C must the equal to 0 individually
Now compare the general equation with the given equation and you will get the value of A, B and C as
\[\begin{array}{l}
A = ({a^2} - 1)\\
B = (a - 1)\\
C = {a^2} - 4a + 3
\end{array}\]
As \[A = 0,B = 0,C = 0\]
\[\begin{array}{l}
\therefore {a^2} - 1 = 0..........................(i)\\
a - 1 = 0................................(ii)\\
{a^2} - 4a + 3 = 0.......................(iv)
\end{array}\]
Solving equation (i) first we get,
\[\begin{array}{l}
\therefore {a^2} - 1 = 0\\
 \Rightarrow (a - 1)(a + 1) = 0\\
 \Rightarrow a = 1, - 1
\end{array}\]
In equation (ii) we get,
\[\begin{array}{l}
 \Rightarrow (a - 1) = 0\\
 \Rightarrow a = 1
\end{array}\]
And finally in equation (iii) we are getting
\[\begin{array}{l}
 \Rightarrow {a^2} - 3a - a + 3 = 0\\
 \Rightarrow a(a - 1) - 3(a - 1) = 0\\
 \Rightarrow (a - 3)(a - 1) = 0\\
 \Rightarrow a = 3,1
\end{array}\]
Clearly by comparing the results we can see that \[a = 1\] is present everywhere.
Therefore \[a = 1\] is the answer.

Note: You can also do the same in a different method just by solving the C part. i.e., \[{a^2} - 4a + 3\] by solving C we are getting \[(a - 3)(a - 1)\] Now replacing the value of C in the actual equation we will get it as
\[({a^2} - 1){x^2} + (a - 1)x + (a - 3)(a - 1) = 0\]
Now it is clearly visible that we can take a \[(a - 1)\] common and we will get the final equation
\[\begin{array}{l}
({a^2} - 1){x^2} + (a - 1)x + (a - 3)(a - 1) = 0\\
 \Rightarrow (a - 1)(a + 1){x^2} + (a - 1)x + (a - 3)(a - 1) = 0\\
 \Rightarrow (a - 1)\left[ {(a + 1){x^2} + x + (a - 3)} \right] = 0\\
\therefore a - 1 = 0\\
 \Rightarrow a = 1
\end{array}\]
So by both the methods we are getting the same result.