Question

Please please please help

Answer 1

II. A. true

I. B. true

Explanation:

we can use the quadratic equation:

`x=\frac{-b+-\sqrt{b^(2)-4*a*c}}{2*a}`

we have:

1)
`x^(2) -4x+5`

a=1 b=-4 c=5

`x=\frac{-(-4)+-\sqrt{(-4)^(2)-4*1*5}}{2*1} \\\\x=(4+-√(-4))/(2) \\\\x=(4+-√(4) i)/(2)\\\\x=(4+-2i)/(2) \\\\x_(1) =2-i\\\\x_(2) =2+i`

the quadratic expression has two complex factors.

2)
`x^(2) -5x+14`

a=1 b=-5 c=14

`x=\frac{-(-5)+-\sqrt{(-5)^(2)-4*1*14}}{2*1} \\\\x=(5+-√(-31))/(2) \\\\x=(5+-√(31) i)/(2)\\\\x_(1) =(5-√(31)i )/(2)\\ \\x_(2) =(5+√(31)i )/(2)`

the solution of
`x^(2) -5x+14` are

`x_(1) =(5-√(31)i )/(2)\\\\or\\x_(2) =(5+√(31)i )/(2)`

3)
`2x^(2) -8x+5`

a=2 b=-8 c=5

`x=\frac{-(-8)+-\sqrt{(-8)^(2)-4*2*5}}{2*2} \\\\x=(8+-√(24))/(4) \\\\x=(8+-2√(6))/(4)\\\\x_(1) =2-(√(6))/(2)\\ \\x_(2) =2+(√(6))/(2)`