# Find the nature of the roots { if it exists by quadratic equation}6x^2 -x-2=0x^2 -2x+1=0x^2+x+1=02x^2 +5x+5=0

2
by aaa345

2014-10-29T20:33:23+05:30
Nature of the root can be identified by D=b^2-4ac
If it is -ve then root will imaginary
If it is +ve then root will be real and different
If it is equal to Zero, then both root will be real and equal
1) 6x^2-x - 2=0
ax^2+bx+c=o
on comparing , we will find D
1-4*6*-2= -ve then root will be imaginary
2) x^2+x+1=0
D= 1 -4*1*1 = -ve,then rootwill be imaginary
3) 2x^2+5x+5=0
D= 25-4*2*5= -ve, root will be imaginary

In the language of complex numbers they are denoted by ω and ω^2
plz send your question on main page, it is difficult to solve this in coulomn
ok sorry
kaushik and jackie there are three question plz solve them
In first question D is +ve, hence root will be real and finite
2014-10-30T13:15:45+05:30

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We use discriminant for this purpose for the equation ax² + bx + c = 0

D = b² - 4ac

If D > 0 roots are real and distinct
D < 0 roots are imaginary and distinct
D = 0 Roots are real and equal

so
1)  6x² - x - 2 = 0
D = (-1²) - 4 x 6 x - 2
D = 1 + 48 = 49
So D > 0 roots are real and distinct

2) x² - 2x + 1 = 0
D = (- 2)
² - 4 x 1 x 1
D = 4 - 4 = 0
So    D = 0 Roots are real and equal

3) x² + x + 1 = 0
D = 1
² - 4 x 1 x 1 = 1 - 4 = - 3
So  D < 0 roots are imaginary and distinct

4) 2x² + 5x + 5 = 0
D = 5
² - 4 x 5 x 2 = 25 - 40 = - 15
So D < 0 roots are imaginary and distinct
hope it helps
plz mark as best