If the sum of the squares of the roots of $x^{2} + p x - 3 = 0$ is $10$ , then the value of $p =$

\pm 2

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\pm 3

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5

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- 5

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Solution

The correct option is A $\pm 2$
$α + β = - p, α β = - 3$
Given $α^{2} + β^{2} = 10$
$\Rightarrow (α + β)^{2} - 2 α β = 10$
$\Rightarrow p^{2} - 2 (- 3) = 10$
$\Rightarrow p^{2} = 4$
$\Rightarrow p = \pm 2$

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