Math problem:how to identify a nature of the roots?

Question

Ok the porlbmei s

1. 2x^2-6x+5=0
2. 2x^2-6x-5=0

It says we have to determine a nature of the roots..

i really dont get it how do you identify a nature of roots,
and also how many natures of roots are there?

Answer 1

Question 1
x = [ 6 ± √(36 - 40) ] / 4
x = [ 6 ± √ (- 4) ] / 4
x = [ 6 ± 2i ] / 4
x = (1/2)[ 3 ± i ]
Imaginary roots.

Question 2
x = [ 6 ± √(36 + 40) ] / 4
x = [ 6 ± √(76) ] / 4
x = [ 6 ± 2√(19) ] / 4
x = (1/2) [ 3 ± √(19) ]
Real and distinct roots

Answer 2

These both equations are quadratic.
So ,if we use shreedharacharya's formula or quadratic formula which is :
[-b +-sq.root{(b^2)-4ac}]/2a
out of which
b^2-4ac is called the discriminant which is denoted by D
and
if D>0 roots are real and distinct
if D<0 roots are not real and are complex
if D=0 then roots are real and equal
Now,
solutions for your questions are;
1 2x^2 -6x +5 =0
now,
b= -6
a= 2
c= 5
So,
D= -6^2 -4*5*2
D= 36-40
D= -4
Since
D<0 so the roots are not real and are complex.

2.
2x^2-6x-5
b= -6
a= 2
c= - 5
So,
D= -6^2 -4*2*-5
D= 36+40
D=76
Since,
D>0
Hence the roots are real and distinct

Answer 3

identify the nature of the roots using the determinant, which is b^2 - 4ac...

b is defined as the coefficient of x,
a is the cooefficient of x^2,
and c is the constant, or a number without any unknown..

if the determinant is

- equal to zero, then u have two real and equal roots
- more than zero, two real but different roots
- less than zero, the eqn have imaginary roots...which means there is no solution for the eqn.

Answer 4

Quadratic formula is

x = [-b +- sqrt(b^2 - 4ac)]/2a

and b^2 - 4ac is the discriminant

If b^2 - 4ac = 0 one {or two equal} root

If b^2 - 4ac > 0 two real roots

If b^2 - 4ac < 0 two imaginary roots

1. 2x^2-6x+5=0

b^2 - 4ac = (-6)^2 - 4(2)(5) = 36 - 40 = - 4
Two imaginary roots

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

b^2 - 4ac = (-6)^2 - 4(2)(-5) = 36 + 40 = 76
Two real roots

Perhaps you were napping when this topic was explained?

Answer 5

Plug it in the discriminant b^2 - 4ac.

1. b^2 - 4ac = 6^2 - 4(2)(5) = -4 < 0

So the roots are both imaginary.

2. b^2 - 4ac = 6^2 - 4(2)(-5) = 66 > 0

So the roots are both real (and irrational).