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]

How do you find the discriminant and how many solutions does 2x^2-7x-4=02x^2-7x-4=0 have?