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Svetllana [295]
3 years ago
14

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

Mathematics
2 answers:
Ganezh [65]3 years ago
6 0
1)B two real roots.
2)D two complex roots.
3) and 4) later
LenKa [72]3 years ago
5 0

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}  )


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