I just wonder:?

Question

How do you get the limit of the quadratic equation as a approaches 0 (that is, the quadratic equation becomes a linear equation)

lim a ->0 [-b ± √(b² - 4ac)]/2a

^_^

Answer 1

Use L'Hopital's Rule. Take the derivative of the expression with respect to a, and the result will give you the limit. You can grind out the math.

Answer 2

I'll assume you mean a->0 from the left. We can use L'Hopital's Rule on the rightmost zero.

lim a->0 d/da[-b + sqrt(b^2 - 4ac)] / d/da[2a]

=

lim a->0 [-4c/(2sqrt(b^2 - 4ac))/2]

= -4c/sqrt(b^2)/4 = -c/b (let's ignore that absolute value...)

As for the other one, it decreases without bound, so there is no limit for it.

Answer 3

b² - 4ac=0

Answer 4

When a is extreme small you get
x = (- b - b)/(2a) This goes to negative infinity.
x = (- b + b)/(2a) goes to 0/0.
The last one you need De l'Hôptal's rule and you have to differentiate with respect to a. To difficult for me.

Answer 5

my memory tells me that the expression you offer :

[-b ± √(b² - 4ac)]/2a

represents "solutions" of the quadratic eqn:

y = ax^2 + bx + c

I don't really understand why you would not
"keep it simple" and just set a=0
and solve the resulting eqn:

y = bx + c

??

rather than trying to apply function-limit concepts to
constant coefficients instead of 'variables'.

good luck

Answer 6

if hte limit tends to 0............. solve the equation,,, if it comes to 0 the the the limit does not exist,,,,,,,, if ur still wit probs jus mail me the Problem ill solve it n rply to u
[e-mail]

Answer 7

-c/b
you have to use the L'Potial rule