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(This question has multiple correct options)

A). -1

B). 2

C). 1

D). 5

Answer

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The given quadratic equation is: ${x^2} + ax + a + 1 = 0$

So, by applying quadratic formula we get

$x = \dfrac{{ - a \pm \sqrt {{a^2} - 4(a + 1)} }}{{2a}}$

= $x = \dfrac{{ - a \pm \sqrt {{a^2} - 4a - 4} }}{{2a}}$

So, to get the integral values of a the expression ${a^2} - 4a - 4$ need to be a perfect square

So ${a^2} - 4a - 4 = {p^2}$

Here ${p^2}$ is a perfect square

\[

{a^2} - 4a - 4 - {p^2} = 0 \\

{a^2} - 4a = 4 + {p^2} \\

{a^2} - 4a + 4 = 4 + 4 + {p^2} \\

\]

Here 4 is added to both left- and right-hand sides

As we know that ${\left( {a \pm b} \right)^2} = {a^2} + {b^2} \pm 2ab$

Therefore ${(a - 2)^2} = 8 + {p^2}$

Let the value of p=1

$

{(a - 2)^2} = 8 + 1 \\

{(a - 2)^2} = 9 \\

a - 2 = \pm 3 \\

$

Let a – 2 = 3

This gives a = 5

And a – 2 = -3

This gives a = - 1

So, the options A and D are correct