Question

1). If the discriminant of a quadratic equation is 28 describe the roots

Answer 1

1)B two real roots.
2)D two complex roots.
3) and 4) later

Answer 2

1. Answer :B. 2 rational roots

Discriminant D >0 -----> 2 rational roots

Discriminant D =0 -----> 1 real root

Discriminant D <0 -----> 2 imaginary roots

D = 28 >0 So 2 rational roots

2. Answer : 2 complex roots

5x^2 -2x+1=0

Discriminant D =$b^2 - 4*a*c= (-2)^2 -4*(5)(1) = 4 - 20 = -16$

D = -16 <0 so 2 complex roots

3. 3x^2 +9=5-4x

$3x^2 +9=5-4x3x^2 +4x +9-5 = 03x^2 + 4x +4 = 0$

Now solve for x, use quadratic formula

$x = \frac{-b+-\sqrt{b^2- 4ac}}{2a}$

$x = \frac{-4+-\sqrt{4^2 - 4*3*4}}{2*3}$

$x = \frac{-4+-\sqrt{-32}}{6}$

x =$\frac{-4 + - 4i\sqrt{2} }{6}$

$x= \frac{-2 + - 2i\sqrt{2} }{3}$

The roots are complex

Solution set is $( \frac{-2 + 2i\sqrt{2} }{3} , \frac{-2 - 2i\sqrt{2} }{3} )$

4. x^2-6x-5=0

$x = \frac{-b+-\sqrt{b^2- 4ac}}{2a}$

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

$x = \frac{6+-\sqrt{56}}{2}$

$x = \frac{6+-2\sqrt{14}}{2}$

$x =3 +-\sqrt{14}$

Solution set ${ 3 +\sqrt{14} , 3 -\sqrt{14} )$