Sure! Let's solve the quadratic equation [tex]\( v^2 + 5v + 4 = 0 \)[/tex] step by step using the quadratic formula.
The general form of a quadratic equation is:
[tex]\[ ax^2 + bx + c = 0 \][/tex]
For the given equation [tex]\( v^2 + 5v + 4 = 0 \)[/tex]:
- [tex]\( a = 1 \)[/tex]
- [tex]\( b = 5 \)[/tex]
- [tex]\( c = 4 \)[/tex]
The quadratic formula is:
[tex]\[ v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
### Step 1: Calculate the Discriminant
The discriminant [tex]\(\Delta\)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = 5^2 - 4 \cdot 1 \cdot 4 \][/tex]
[tex]\[ \Delta = 25 - 16 \][/tex]
[tex]\[ \Delta = 9 \][/tex]
### Step 2: Find the Square Root of the Discriminant
[tex]\[ \sqrt{\Delta} = \sqrt{9} = 3 \][/tex]
### Step 3: Compute the Solutions
Using the quadratic formula, we now solve for [tex]\( v \)[/tex]:
#### First Solution:
[tex]\[ v_1 = \frac{-b + \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ v_1 = \frac{-(5) + 3}{2 \cdot 1} \][/tex]
[tex]\[ v_1 = \frac{-5 + 3}{2} \][/tex]
[tex]\[ v_1 = \frac{-2}{2} \][/tex]
[tex]\[ v_1 = -1 \][/tex]
#### Second Solution:
[tex]\[ v_2 = \frac{-b - \sqrt{\Delta}}{2a} \][/tex]
[tex]\[ v_2 = \frac{-(5) - 3}{2 \cdot 1} \][/tex]
[tex]\[ v_2 = \frac{-5 - 3}{2} \][/tex]
[tex]\[ v_2 = \frac{-8}{2} \][/tex]
[tex]\[ v_2 = -4 \][/tex]
### Conclusion
The solutions to the quadratic equation [tex]\( v^2 + 5v + 4 = 0 \)[/tex] are:
[tex]\[ v_1 = -1 \][/tex]
[tex]\[ v_2 = -4 \][/tex]
Additionally, the discriminant [tex]\(\Delta\)[/tex] was found to be 9, confirming that the equation has two distinct real roots.
