Question

Please please please help

Answer 1

Answer:

I. B. true

II. A true

Step-by-step 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

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

the quadratic expression has two complex factors.

2)$x^{2} -5x+14$

a=1 b=-5 c=14

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

the solution of $x^{2} -5x+14$ are

$x_{1} =\frac{5-\sqrt{31}i }{2}\\\\or\\x_{2} =\frac{5+\sqrt{31}i }{2}$

3)$2x^{2} -8x+5$

a=2 b=-8 c=5

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