compute the value of the discriminant and give the number of real solutions?

Question

compute the value of the discriminant and give the number of real solutions to the quadraic equation

5x^2+x+2=0

Discriminant:

Number of REAL solutions

there is another one with the need of the same answers:

4x^2 + 4x + 1 = 0

discriminant:
Number of REal Solutions

I still do not understand which is the disccriminant for the first equation
is it -39?
with 0 solutions

The second was -1/10
with 0 real solutions

Answer 1

The first answer is not exactly right. The roots of the quadratic are
(-b(+/-)sqrt(b^2-4*a*c)) / (2*a) (not -b^2...)
However the solution is correct; the discriminant is b^2-4*a*c which for the first quadratic is 1^2 - 4*5*2 = -39. Per the ref., a negative discriminant means both roots are complex. They are: (-1(+/-)sqrt(39)*i ) / 10.
For the second quadratic, the discriminant is 4*4 - 4*4*1 = 0. As a result there is one real root: -4 / 8 or -0.5.
Re your followup question, the summary is:
1st equation, disc. = -39, there are no real roots and 2 complex roots,
(-1+sqrt(39)*i ) / 10 and (-1-sqrt(39)*i ) / 10.
2nd equation, disc. = 0, there is one real root (0.5) and no complex roots.

Answer 2

the discriminant is of the form: b^2-4ac. In this case the discriminant is -39. This means your solutions will be imaginary and you have to use the quadratic formula. That is of the form: -b^2(+/-)sqrt(discriminant)/2a. those values would be;

-1/10 (+/-) sqrt(39)i/10

sqrt being square root and (+/-) meaning there are two solutions, one the sum of the two values and the other the difference.


No real solutions