Question

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

A. 1 rational root
B. 2 rational roots
C. 2 irrational roots
D. 2 complex roots


2). Describe the roots of 5x^2 -2x+1=0

A. 1 rational root
B. 2 rational roots
C. 2 irrational roots
D. 2 complex roots

3. What is the solution set for 3x^2 +9=5-4x?

4. What is the solution set for x^2-6x-5=0

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 = (-b+-√(b^2- 4ac))/(2a)`

`x = (-4+-√(4^2 - 4*3*4))/(2*3)`

`x = (-4+-√(-32))/(6)`

x =
`(-4 + - 4i√(2) )/(6)`

`x= (-2 + - 2i√(2) )/(3)`

The roots are complex

Solution set is
`( (-2 + 2i√(2) )/(3) , (-2 - 2i√(2) )/(3) )`

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

`x = (-b+-√(b^2- 4ac))/(2a)`

`x = (-(-6)+-√((-6)^2- 4*1*(-5)))/(2*1)`

`x = (6+-√(56))/(2)`

`x = (6+-2√(14))/(2)`

`x =3 +-√(14)`

Solution set
`{ 3 +√(14) , 3 -√(14) )`