Question

Hii guys!!!

Find the nature of the roots of the quadratic equation 6x^2+x-2=0, if real exist find them.

Answer 1

Equation:

6x² + x - 2 = 0

Determinant , D = b² - 4ac

==> b² - 4ac

On comparing,

a = 6 ; b = 1 ; c = -2

==> 1² - 4×6×-2

==> 1 + 4×6×2

==> 1 + 48

==> 49

Since, D > 0 ,So real and distinct roots exist.

Now,

6x² + x - 2 = 0

Also,

a = 6 ; b = 1 ; c = -2

x = -b ± √D / 2a

x = -1 ± √49 / 12

Case-1

x = -1 + 7 / 12

x = 6/12

x = 1/2

Case-2

x = -1 - 7 / 12

x = -8/12

x = -2/3

Hence, Zeroes are 1/2 and -2/3