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Marat540 [252]
4 years ago
14

Please please please help

Mathematics
1 answer:
Mrrafil [7]4 years ago
7 0

Answer:

II. A. true

I. B. 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}

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