To solve the quadratic equation [tex]\( x^2 - 5x - 14 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
First, let's identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the quadratic equation in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = -5 \)[/tex]
- [tex]\( c = -14 \)[/tex]
Next, we need to calculate the discriminant. The discriminant is given by the formula:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-5)^2 - 4 \cdot 1 \cdot (-14) \][/tex]
[tex]\[ \Delta = 25 + 56 \][/tex]
[tex]\[ \Delta = 81 \][/tex]
The discriminant is [tex]\( 81 \)[/tex], which is a positive number, indicating that there are two distinct real solutions.
Now, apply the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Since [tex]\(\Delta = 81 \)[/tex], [tex]\(\sqrt{\Delta} = 9\)[/tex].
Substitute [tex]\( b \)[/tex], [tex]\( \sqrt{\Delta} \)[/tex], and [tex]\( a \)[/tex] into the formula:
[tex]\[ x = \frac{5 \pm 9}{2 \cdot 1} \][/tex]
This gives us two solutions:
[tex]\[ x_1 = \frac{5 + 9}{2} = \frac{14}{2} = 7 \][/tex]
[tex]\[ x_2 = \frac{5 - 9}{2} = \frac{-4}{2} = -2 \][/tex]
Thus, the solutions to the quadratic equation [tex]\( x^2 - 5x - 14 = 0 \)[/tex] are:
- Lesser [tex]\( x = -2 \)[/tex]
- Greater [tex]\( x = 7 \)[/tex]
So, the answers are:
[tex]\[
\text{lesser } x = -2
\][/tex]
[tex]\[
\text{greater } x = 7
\][/tex]
