Question

Joline is solving the equation 0 = x2 – 5x – 4 using the quadratic formula. Which value is the negative real number solution to her quadratic equation? Round to the nearest tenth if necessary. Quadratic formula: x = StartFraction negative b plus or minus StartRoot b squared minus 4 a c EndRoot Over 2 a EndFraction –5.7 –4 –1 –0.7

Answer 1

Answer:

- 0.7

Step-by-step explanation:

We are given the general quadratic formula, as " $x = -b \frac{+}{-} \sqrt{(b^2-4ac)} / 2a$. " 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 ) $x = - b+ ( \sqrt{b^2-4ac} ) / 2a$,

2 ) $x = - b- ( \sqrt{b^2-4ac} ) / 2a$

We know that a = 1, b = - 5, and c = - 4 from the equation " $0 = x^2 - 5x - 4$ " ( which can be rewritten in the form $0 = ax^2 + bx + c$ ). Therefore, simply plug in these values into the quadratic formula to receive two solutions:

$x = 5 + ( \sqrt{( - 5 )^2 - 4( 1 )( - 4 )} ) / 2( 1 ),\\x = 5 + ( \sqrt{25 + 16} ) / 2,\\x = 5 + \sqrt{41} / 2,\\x = ( About ) 5.7$- This is our positive solution

__________

$x = 5 - ( \sqrt{( - 5 )^2 - 4( 1 )( - 4 )} ) / 2( 1 ),\\x = 5 - ( \sqrt{25 + 16} ) / 2,\\x = 5 - \sqrt{41} / 2,\\x = ( About ) - 0.7$- And this is our negative solution

You can see that - 0.7 is the negative real number solution to Joline's quadratic equation!