1. Determine the discriminant and nature of roots of each quadratic equation. a.) x2 -6x + 9=0
b.) x2 - 4x +3=0
c.) x2 -7x - 4=0
d.) 2x2 + 3x + 5=0

Please check the explanation.

Step-by-step explanation:

The general quadratic equation is

ax²+bx+c=0

The discriminant = D = b² - 4ac

When b² - 4ac = 0 there is one real root.

When b² - 4ac > 0 there are two real roots.

When b² - 4ac < 0 there are two complex roots.

1) x² -6x + 9=0

On comparing with given quadratic equation x² -6x + 9=0

a = 1, b=-6, c=9

D = b² - 4ac

   = (-6)² - 4(1)(9)

   = 36 - 36

   = 0

D = 0

Thus, there is one real root of quadratic equation x² -6x + 9=0.

2) x² -4x + 3=0

On comparing with given quadratic equation x² -4x + 3=0

a = 1, b=-4, c=3

D = b² - 4ac

   = (-4)² - 4(1)(3)

   = 16 - 12

   = 4

D > 0

Thus, there are two real roots of quadratic equation x² -4x + 3=0.

3)  x² -7x - 4=0

On comparing with given quadratic equation x² -7x - 4=0

a = 1, b=-7, c=-4

D = b² - 4ac

   = (-7)² - 4(1)(-4)

   = 49 + 16

   = 65

D > 0

Thus, there are two real roots of quadratic equation x² -7x - 4=0

4)  2x² +3x +5=0

On comparing with given quadratic equation 2x² +3x +5=0

a = 2, b=3, c=5

D = b² - 4ac

   = (3)² - 4(2)(5)

   = 9 - 40

   = -31

D < 0

Thus, there are two complex roots of quadratic equation 2x² +3x +5=0

its 69.420

step-by-step explanation: