Let's solve the quadratic equation [tex]\( 15t^2 + 7t - 4 = 0 \)[/tex] using the quadratic formula.
### Part (A):
The quadratic formula is given by:
[tex]\[ t = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
In this equation:
- [tex]\( a = 15 \)[/tex]
- [tex]\( b = 7 \)[/tex]
- [tex]\( c = -4 \)[/tex]
First, define the discriminant [tex]\( D \)[/tex]:
[tex]\[ D = b^2 - 4ac \][/tex]
[tex]\[ D = 7^2 - 4 \cdot 15 \cdot (-4) \][/tex]
[tex]\[ D = 49 + 240 \][/tex]
[tex]\[ D = 289 \][/tex]
The formula then becomes:
[tex]\[ t = \frac{-b \pm \sqrt{D}}{2a} \][/tex]
Let’s find the appropriate values of [tex]\( N \)[/tex], [tex]\( M \)[/tex], and [tex]\( D \)[/tex]:
- [tex]\( N = -b = -7 \)[/tex]
- [tex]\( M = 2a = 2 \cdot 15 = 30 \)[/tex]
- [tex]\( D = 289 \)[/tex]
So, the integer values are:
[tex]\[ N = -7 \][/tex]
[tex]\[ M = 30 \][/tex]
[tex]\[ D = 289 \][/tex]
### Part (B):
Now let's find the actual solutions by simplifying the radical and solving for [tex]\( t \)[/tex].
The quadratic formula in this case is:
[tex]\[ t = \frac{-7 \pm \sqrt{289}}{30} \][/tex]
Since [tex]\( \sqrt{289} = 17 \)[/tex], the equation becomes:
[tex]\[ t = \frac{-7 \pm 17}{30} \][/tex]
This gives us two solutions:
1. [tex]\( t_1 = \frac{-7 + 17}{30} = \frac{10}{30} = \frac{1}{3} \)[/tex]
2. [tex]\( t_2 = \frac{-7 - 17}{30} = \frac{-24}{30} = -\frac{4}{5} \)[/tex]
Therefore, the simplified solutions are:
[tex]\[ t = \frac{1}{3}, -\frac{4}{5} \][/tex]
So,
[tex]\[ t = \frac{1}{3}, -\frac{4}{5} \][/tex]
