How do you find the discriminant and how many solutions does #2x^2-7x-4=0# have?

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Answer 1 · 2018-07-09T10:43:29

Discriminant #=81# & equation has #2# solutions i.e. two real distinct roots

The given quadratic equation: #2x^2-7x-4=0# will always have two solutions .

Now, comparing with #ax^2+bx+c=0#, we get

#a=2, b=-7# & #c=-4#

The discriminant #\Delta=b^2-4ac# is given as

#b^2-4ac=(-7)^2-4(2)(-4)=81>0#

The given quadratic equation has two real distinct roots.

Answer 2 · 2018-07-09T10:53:35

Two Solutions: #x=4 and x= -0.5 #

# 2 x^2-7 x-4=0#

Comparing with standard quadratic equation #ax^2+bx+c=0#

# a=2 ,b=-7 , c=- 4# Discriminant # D= b^2-4a c#

#D=49+32 =81#, discriminant is positive, we get two real

solutions. Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (7+-sqrt 81)/4 = (7+-9)/4:. x =(7+9)/4= 16/4=4 # and

#x =(7-9)/4= -2/4= -0.5:. x=4 and x= -0.5 #

Solution: #x=4 and x= -0.5 # [Ans]