Question 48947
-6x^2-5x+1=0...STD.EQN.IS 
AX^2+BX+C=0..................DISCRIMINANT =B^2-4AC
HERE WE HAVE A=-6...B=-5....C=1...OK?
compute the value of the deiscriminant and give the number of real solutions of the quadratic equation.
D=(-5)^2-4(-6)(1)=25+24=49...+VE & PERFECT SQUARE......HENCE REAL ,RATIONAL SOLUTION WITH 2 DISTINCT ROOTS.
-6x^2-5x+1=0
compute the value of the deiscriminant and give the number of real solutions of the quadratic equation.
D=(-5)^2-4(-6)(1)=25+24=49...+VE & PERFECT SQUARE......HENCE REAL ,RATIONAL SOLUTION WITH 2 DISTINCT ROOTS.