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To find the zero(s) of the function [tex]\( f(x) = x^2 - 2x - 4 \)[/tex], we need to determine the values of [tex]\( x \)[/tex] that make [tex]\( f(x) = 0 \)[/tex]. This involves solving the quadratic equation:
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Here is a detailed, step-by-step solution to solve the quadratic equation:
1. Identify the coefficients:
The quadratic equation is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. For our equation [tex]\( x^2 - 2x - 4 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
- [tex]\( c = -4 \)[/tex]
2. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
3. Plug in the coefficients:
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -4 \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
Simplifying this:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{20}}{2} \][/tex]
4. Simplify further:
Simplify the expression inside the square root:
[tex]\[ x = \frac{2 \pm \sqrt{4 \cdot 5}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 2 \sqrt{5}}{2} \][/tex]
5. Factor out common terms:
Divide both the numerator and the denominator by 2:
[tex]\[ x = 1 \pm \sqrt{5} \][/tex]
6. Write the solutions:
Thus, the solutions to the equation [tex]\( x^2 - 2x - 4 = 0 \)[/tex] are:
[tex]\[ x = 1 - \sqrt{5} \quad \text{and} \quad x = 1 + \sqrt{5} \][/tex]
Therefore, the zero(s) of the function [tex]\( f(x) = x^2 - 2x - 4 \)[/tex] are:
[tex]\[ x = 1 - \sqrt{5} \quad \text{and} \quad x = 1 + \sqrt{5} \][/tex]
[tex]\[ x^2 - 2x - 4 = 0 \][/tex]
Here is a detailed, step-by-step solution to solve the quadratic equation:
1. Identify the coefficients:
The quadratic equation is in the form [tex]\( ax^2 + bx + c = 0 \)[/tex]. For our equation [tex]\( x^2 - 2x - 4 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
- [tex]\( c = -4 \)[/tex]
2. Use the quadratic formula:
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
3. Plug in the coefficients:
Substitute [tex]\( a = 1 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -4 \)[/tex] into the quadratic formula:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \][/tex]
Simplifying this:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 16}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{20}}{2} \][/tex]
4. Simplify further:
Simplify the expression inside the square root:
[tex]\[ x = \frac{2 \pm \sqrt{4 \cdot 5}}{2} \][/tex]
[tex]\[ x = \frac{2 \pm 2 \sqrt{5}}{2} \][/tex]
5. Factor out common terms:
Divide both the numerator and the denominator by 2:
[tex]\[ x = 1 \pm \sqrt{5} \][/tex]
6. Write the solutions:
Thus, the solutions to the equation [tex]\( x^2 - 2x - 4 = 0 \)[/tex] are:
[tex]\[ x = 1 - \sqrt{5} \quad \text{and} \quad x = 1 + \sqrt{5} \][/tex]
Therefore, the zero(s) of the function [tex]\( f(x) = x^2 - 2x - 4 \)[/tex] are:
[tex]\[ x = 1 - \sqrt{5} \quad \text{and} \quad x = 1 + \sqrt{5} \][/tex]
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