Asked by toni
help me please im confused
Consider the quadratic expression X^2 + 4x + c = 0.
For what range of values does the equation have two complex roots?
what am i doing here can someone show me step by step so i can at least try to do the other problems i have
Consider the quadratic expression X^2 + 4x + c = 0.
For what range of values does the equation have two complex roots?
what am i doing here can someone show me step by step so i can at least try to do the other problems i have
Answers
Answered by
Jai
Hi. I answered you previous post and it was almost similar, except that here, the roots must be complex/imaginary so D must be less than zero. Anyway,
Recall the formula for discriminant. For a quadratic equation in the general form, ax^2 + bx + c = 0,
D = b^2 - 4ac
if
D = 0 : real, equal/double root
D > 0 : two real, unequal roots
D < 0 : two imaginary roots
Since we're required to have complex/imaginary, D < 0, and solve for the unknown, c.
x^2 + 4x + c = 0
a = 1
b = 4
c = ?
Substituting to the discriminant formula, (D < 0)
0 < 4^2 - 4*1*c
0 < 16 - 4c
4c < 16
c < 4
Hope this helps~ :3
Recall the formula for discriminant. For a quadratic equation in the general form, ax^2 + bx + c = 0,
D = b^2 - 4ac
if
D = 0 : real, equal/double root
D > 0 : two real, unequal roots
D < 0 : two imaginary roots
Since we're required to have complex/imaginary, D < 0, and solve for the unknown, c.
x^2 + 4x + c = 0
a = 1
b = 4
c = ?
Substituting to the discriminant formula, (D < 0)
0 < 4^2 - 4*1*c
0 < 16 - 4c
4c < 16
c < 4
Hope this helps~ :3
Answered by
Swaastikaa
I disagree that the answer is c < 4. The correct answer is c > 4.
Lets see the steps again:
x^2 + 4x + c = 0, b^2 - 4ac < 0
4^2 - 4(1)(c) < 0
16 - 4c < 0
16 < 4c
4c > 16
Therefore c > 4
Lets see the steps again:
x^2 + 4x + c = 0, b^2 - 4ac < 0
4^2 - 4(1)(c) < 0
16 - 4c < 0
16 < 4c
4c > 16
Therefore c > 4
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