Question

If there are two real solutions to the equation x^2 + 4x + c=0 , which is a possible value of c?



20


17


-5


16

Answer 1

@Hero

Answer 2

@imqwerty

Answer 3

for 2 real solutions the discriminant should be a perfect square :)

Answer 4

so 16?

Answer 5

can u tell what is the value of discriminant when u put c=16

Answer 6

-48?

Answer 7

how

Answer 8

well to have 2 real solution, then the discriminant is

\[b^2 - 4ac \ge 0\]

and the result is either 0 or a square number...

so you have a = 1, b = 4 and c

then

\[4^2- 4 \times 1 \times c \ge 0\]

then

\[16 - 4c \ge 0 \]

so solve the inequality

then find the solution that meets the condition of the inequality.

don't forget, dividing by a negative requires the inequality to be reversed.

and just a quick revision on the discriminant

\[b^2 - 4ac = 0\]

there are 2 equal roots

\[b^2 - 4ac >0 ~~and~~b^2 - 4ac ~~is~a~\square ~number\]

then the roots, are real, unequal and rational.

Hope it makes sense