One root of the quadratic equation $x^{2} + p x + 3 = 0$ is $1$ . Its other root is

- 3

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0

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3

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4

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Solution

The correct option is C $3$
$1$ is the root of the given quadratic equation. Thus,
$(1)^{2} + p (1) + 3 = 0$
$p + 4 = 0$
$p = - 4$
Thus, the equation becomes $x^{2} - 4 x + 3 = 0$
$x^{2} - x - 3 x + 3 = 0$
$x (x - 1) - 3 (x - 1) = 0$
$(x - 1) (x - 3) = 0$
$x = 1, 3$
Hence, the other root of the equation is 3.

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