To solve the quadratic equation [tex]\(0 = x^2 - 2x - 3\)[/tex] using the quadratic formula, we first identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -2 \)[/tex]
- [tex]\( c = -3 \)[/tex]
We use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's substitute the identified values into the formula step-by-step:
1. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the formula:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
2. Simplify the terms:
- The negation of [tex]\( -2 \)[/tex] is [tex]\( 2 \)[/tex]:
[tex]\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
- Calculate the square of [tex]\( -2 \)[/tex]:
[tex]\[ (-2)^2 = 4 \][/tex]
- Calculate the product [tex]\( 4 \cdot 1 \cdot (-3) \)[/tex]:
[tex]\[ 4 \cdot -3 = -12 \][/tex]
- The quadratic formula under the square root now simplifies to:
[tex]\[ x = \frac{2 \pm \sqrt{4 + 12}}{2 \cdot 1} \][/tex]
3. Add the numbers inside the square root:
[tex]\[ x = \frac{2 \pm \sqrt{16}}{2} \][/tex]
4. Calculate the square root of 16:
[tex]\[ \sqrt{16} = 4 \][/tex]
5. Substitute this value back into the formula:
[tex]\[ x = \frac{2 \pm 4}{2} \][/tex]
6. Solve for the two possible values of [tex]\( x \)[/tex]:
- When adding 4:
[tex]\[ x = \frac{2 + 4}{2} = \frac{6}{2} = 3 \][/tex]
- When subtracting 4:
[tex]\[ x = \frac{2 - 4}{2} = \frac{-2}{2} = -1 \][/tex]
Therefore, the correct substitution of the values [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] into the quadratic formula is:
[tex]\[ \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \][/tex]
And the final solutions to the quadratic equation [tex]\(0 = x^2 - 2x - 3\)[/tex] are:
[tex]\[ x = 3 \][/tex]
[tex]\[ x = -1 \][/tex]
Thus, the answer matches:
[tex]\[ \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2(1)} \][/tex]
