The zeros of the equation $x^{2} + 8 x + 9 = 0$ , can be

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x = - 1 \pm \sqrt{7}

x = - 2 \pm \sqrt{7}

x = 4 \pm \sqrt{7}

x = - 4 \pm \sqrt{7}

answer is D.

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Detailed Solution

It is given that the equation is

x^{2} + 8 x + 9 = 0 .

Here we use the quadratic formula to solve the question i.e.,

\frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

.
Comparing the values of the given equation with standard quadratic equation,

{ax}^{2} + bx + c

we get the values of a, b, c as

a = 1, b = 8, c = 9

.
Substituting the given values in the quadratic formula,

x = \frac{- b \pm \sqrt{b^{2} - 4 ac}}{2 a}

x = \frac{- 8 \pm \sqrt{64 - 4 (1) (9)}}{2 \times 1}

x = \frac{- 8 \pm \sqrt{28}}{2}

x = \frac{- 8 \pm 2 \sqrt{7}}{2}

∴ x = - 4 \pm \sqrt{7}

∴

Zeros of the given equation are

- 4 \pm \sqrt{7}

.
Hence, option 4.

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