The condition that $x^{4} + a x^{3} + b x^{2} + c x + d$ is a perfect square, is:

c^{2} = a d

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c^{2} = a d^{2}

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c^{2} = a^{2} d^{2}

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c^{2} = a^{2} d

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Solution

The correct option is A $c^{2} = a d$
Solution :-

Let - $x^{4} + a x^{3} + b x^{2} + c x + d = (x^{2} + p x + q)^{2}$

$= x^{4} + p^{2} x^{2} + q^{2} + 2 x^{3} p + 2 p q x + 2 x^{2} q$

Now comparing the coefficient of $x^{3}, x^{2}, x$ and constant

we get $a = 2 p . . . . (1)$

$b = p^{2} + 2 q . . . . (2)$

$c = 2 p q . . . . (3)$

$d = q^{2} . . . . (4)$

we get $p = \frac{a}{2}$ and $q = \frac{c}{2 p} = \frac{c}{2 p}$ by $(1)$ & $(3)$

Substituting $p$ & $q$ in $(2)$ & $(4)$

$b = \frac{a^{2}}{4} + \frac{2 c}{a}$

$d = \frac{c^{2}}{a^{2}} \Rightarrow c^{2} = a d$

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