The value of $a$ for which the $(a^{2} - 4) x^{2} + a^{2} - 5 a + 6 = 0$ is an identity in $x$ is

1

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Solution

The correct option is C $2$
Given $(a^{2} - 4) x^{2} + a^{2} - 5 a + 6 = 0$
For this to be an identity in $x$ , the coefficients of various powers of $x$ and constant term must be zero.

Put coefficient of $x^{2} = 0$ ,
$\Rightarrow (a^{2} - 4) = 0$
$\Rightarrow (a + 2) (a - 2) = 0$
$\Rightarrow a = - 2, 2$

Put constant term $= 0$ ,
$\Rightarrow a^{2} - 5 a + 6 = 0$
$\Rightarrow (a - 2) (a - 3) = 0$
$\Rightarrow a = 2, 3$

Hence, the common value of $a$ is $2$ .
$\Rightarrow a = 2$

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The value of a for which the (a2−4)x2+a2−5a+6=0 is an identity in x is

The value of $a$ for which the $(a^{2} - 4) x^{2} + a^{2} - 5 a + 6 = 0$ is an identity in $x$ is