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

D

Explanation:

the letter to the answer -0.7

Answer 2

- 0.7

Explanation:

We are given the general quadratic formula, as "
`x = -b (+)/(-) √((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+ ( √(b^2-4ac) ) / 2a`,

2 )
`x = - b- ( √(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 + ( √(( - 5 )^2 - 4( 1 )( - 4 )) ) / 2( 1 ),\\x = 5 + ( √(25 + 16) ) / 2,\\x = 5 + √(41) / 2,\\x = ( About ) 5.7`- This is our positive solution

__________

`x = 5 - ( √(( - 5 )^2 - 4( 1 )( - 4 )) ) / 2( 1 ),\\x = 5 - ( √(25 + 16) ) / 2,\\x = 5 - √(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!