Let's find the zeros of the quadratic function [tex]\( x^2 - 4x - 5 \)[/tex].
To find the zeros (or roots) of a quadratic function, we use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
- The coefficient [tex]\( a \)[/tex] is 1 (from [tex]\( x^2 \)[/tex]),
- The coefficient [tex]\( b \)[/tex] is -4 (from [tex]\( -4x \)[/tex]),
- The constant [tex]\( c \)[/tex] is -5.
First, we calculate the discriminant [tex]\( \Delta \)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-4)^2 - 4(1)(-5) \][/tex]
[tex]\[ \Delta = 16 + 20 \][/tex]
[tex]\[ \Delta = 36 \][/tex]
Since the discriminant is positive, there are two real roots. We can now find the roots using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\( \Delta = 36 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( a = 1 \)[/tex]:
[tex]\[ x = \frac{4 \pm \sqrt{36}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm 6}{2} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{4 + 6}{2} = \frac{10}{2} = 5 \][/tex]
[tex]\[ x_2 = \frac{4 - 6}{2} = \frac{-2}{2} = -1 \][/tex]
Thus, the zeros of the function [tex]\( x^2 - 4x - 5 \)[/tex] are:
[tex]\[ x = 5 \text{ and } x = -1 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{x = -1 \text{ and } x = 5} \][/tex]
