The zeros for the given equation are A. [tex]x = {2 \pm \sqrt{13}}.[/tex]
Step-by-step explanation:
Step 1:
To solve for x in a polynomial equation, we use the formula
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}.[/tex]
Here a is the coefficient of [tex]x^{2}[/tex], b is the coefficient of [tex]x[/tex] and c is the coefficient of the constant term.
In the given equation, [tex]a = 1, b = -4, c = -9[/tex].
Step 2:
Substituting the values of a, b, and c in the equation, we get
[tex]x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} =\frac{-(-4) \pm \sqrt{(-4)^{2}-4 (1)(-9)}}{2 (1)} = \frac{4 \pm \sqrt{(16+36}}{2 }.[/tex]
[tex]\frac{4 \pm \sqrt{(16+36}}{2 } = \frac{4 \pm \sqrt{(52}}{2 } = \frac{4 \pm2 \sqrt{(13}}{2 }= {2 \pm \sqrt{13}}.[/tex]
So the answer is option A. [tex]x = {2 \pm \sqrt{13}}.[/tex]
