SOLUTION: x^2+4x+c=0 what value of c does the equation have one real root and what range of value does it have two roots, fo what range of values does the equation have two complex.

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Question 766331: x^2+4x+c=0
what value of c does the equation have one real root and what range of value does it have two roots, fo what range of values does the equation have two complex.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The general solution, x=%28-4%2B-sqrt%284%2A4-4%2A1%2Ac%29%29%2F%282%2A1%29
contains an expression, 4%5E2-4c.

That is the DISCRIMINANT. It is written in the square root function.

Discriminant = 0, then equation has one real root.
Discriminant greater than zero, then equation has two real roots.
Discriminant less than zero, then the equation has two complex roots with imaginaries.

x=-2%2B-sqrt%284-c%29 when simplified.

c=4 makes x be exactly one solution because sqrt%284-4%29=0.
c<4 makes x be either of two real solutions because sqrt%284-c%29 will itself be a real value.
c>4 makes sqrt%284-c%29 be imaginary, and there be two complex roots solutions for x with imaginary components.