How many real roots are possible for the below equation?
$(x + 1) (x + 2) (x + 3) (x + 4) = 24$

0, 1, 2, 3

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0, 5

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0, 3, 2, -1

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0, -5

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Solution

The correct option is B 0, 5
$[(x + 1) (x + 4)] [(x + 2) (x + 3)] = 24$

$(x^{2} + 5 x + 4) (x^{2} + 5 x + 6) = 24$

Let $x^{2} + 5 x = k$

$(k + 4) (k + 6) = 24$

$k^{2} + 10 k = 0$

$k (k + 10) = 0$

$k = 0, k = - 10$

$x^{2} + 5 x = 0, x^{2} + 5 x = - 10$

$x (x + 5) = 0, x^{2} + 5 x + 10 = 0$

$x = 0, x = - 5; D = 25 - 40 = - 15$ No real roots.

So, roots are 0, -5.

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