Question

show that the equation 2x^2+x-6=0 has real number and solve it (D=b^2-4ac) with alpha and beta​

Answer 1

Answer:

2x²+x-6=0

2x²+4x-3x-6=0

2x(x+2)-3(x+2)=0

(2x-3)(x+2)=0

x=3/2, -2.

hence it has real roots.