Answer:
- 0.7
Step-by-step explanation:
We are given the general quadratic formula, as " [tex]x = -b \frac{+}{-} \sqrt{(b^2-4ac)} / 2a[/tex]. " This can be represented by two separate formula's, which will come in handy when determining the positive and negative roots of the equation ( and here we need the negative root ):
1 ) [tex]x = - b+ ( \sqrt{b^2-4ac} ) / 2a[/tex],
2 ) [tex]x = - b- ( \sqrt{b^2-4ac} ) / 2a[/tex]
We know that a = 1, b = - 5, and c = - 4 from the equation " [tex]0 = x^2 - 5x - 4[/tex] " ( which can be rewritten in the form [tex]0 = ax^2 + bx + c[/tex] ). Therefore, simply plug in these values into the quadratic formula to receive two solutions:
[tex]x = 5 + ( \sqrt{( - 5 )^2 - 4( 1 )( - 4 )} ) / 2( 1 ),\\x = 5 + ( \sqrt{25 + 16} ) / 2,\\x = 5 + \sqrt{41} / 2,\\x = ( About ) 5.7[/tex]- This is our positive solution
__________
[tex]x = 5 - ( \sqrt{( - 5 )^2 - 4( 1 )( - 4 )} ) / 2( 1 ),\\x = 5 - ( \sqrt{25 + 16} ) / 2,\\x = 5 - \sqrt{41} / 2,\\x = ( About ) - 0.7[/tex]- And this is our negative solution
You can see that - 0.7 is the negative real number solution to Joline's quadratic equation!
