The quadratic equation $x^{2} + b x + 4 = 0$ will have real roots if

b \leq - 4

only

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b \geq 4

only

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- 4 < b < 4

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b \leq - 4, b \geq 4

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Solution

The correct option is C $b \leq - 4, b \geq 4$
In any quadratic equation, real roots exist if discriminant is greater than or equal to zero.
So for $x^{2} + b x + 4 = 0$
Discriminant is $= b^{2} - 4 \times 1 \times 4$
So, $b^{2} - 16 \geq 0$
$\Rightarrow b \geq 4, b \leq - 4$

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