Answer:
B
Step-by-step explanation:
Using the quadratic formula with
a = 16, b = 0 and c = - 80, then
x = ( 0 ± [tex]\sqrt{0-(4(16)(-80)}[/tex] / 32
= ± [tex]\frac{\sqrt{5120} }{32}[/tex]
x = - [tex]\frac{\sqrt{5120} }{32}[/tex] , x = [tex]\frac{\sqrt{5120} }{32}[/tex]
x = - 2.24, x = 2.24 ( to the nearest hundredth )
For this case we have the following quadratic equation:
[tex]16x ^ 2-80 = 0[/tex]
Applying the quadratic formula we have:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 16\\b = 0\\c = -80[/tex]
Substituting we have:
[tex]x = \frac {-0 \pm \sqrt {0 ^ 2-4 (16) (- 80)}} {2 (16)}\\x = \frac {\pm \sqrt {4 (16) (80)}} {32}\\x = \frac {\pm \sqrt {5120}} {32}\\x = \frac {\pm71.55} {32}[/tex]
We have two roots:[tex]x_ {1} = \frac {-71.55} {32} = - 2.24\\x_ {2} = \frac {71.55} {32} = 2.24[/tex]
Answer:
Option B
