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Find the nature of the roots { if it exists by quadratic equation}

6x^2 -x-2=0

x^2 -2x+1=0

x^2+x+1=0

2x^2 +5x+5=0

2
by aaa345

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6x^2 -x-2=0

x^2 -2x+1=0

x^2+x+1=0

2x^2 +5x+5=0

by aaa345

Log in to add a comment

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

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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 = 1² - 4 x 1 x 1 = 1 - 4 = - 3

So D < 0 roots are imaginary and distinct

4) 2x² + 5x + 5 = 0

D = 5² - 4 x 5 x 2 = 25 - 40 = - 15

So D < 0 roots are imaginary and distinct

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 = 1² - 4 x 1 x 1 = 1 - 4 = - 3

So D < 0 roots are imaginary and distinct

4) 2x² + 5x + 5 = 0

D = 5² - 4 x 5 x 2 = 25 - 40 = - 15

So D < 0 roots are imaginary and distinct