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Qua Eqn

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Qua Eqn

by sanju09 » Fri Feb 27, 2009 5:57 am
For what values of 'm' is y = 0, if y = x^2 + (2 m + 1) x + m^2 - 1? x is a real number.

(A) m (= or >) -2
(B) m < 0
(C) m = 0
(D) m (= or >) -1.25
(E) m = -1/5
Last edited by sanju09 on Sat Feb 28, 2009 2:52 am, edited 1 time in total.
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by DanaJ » Fri Feb 27, 2009 6:48 am
What you are really aking for is to know for what m does the given equation have real solutions. This only happens when the discriminant is equal to or greater than 0, meaning you need to find out whether b^2 - 4ac >=0.

Translate this to the present equation and you get:
discriminant = (2m + 1)^2 - 4(m^2 - 1) = 4m^2 + 4m + 1 - 4m^2 + 4 = 4m + 5. This has to be greater than or equal to zero: 4m + 5 >= 0
4m >= -5
m >= -5/4 or m >= -1.25

Answer D
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by Baldini » Wed Mar 04, 2009 3:42 am
Hi DanaJ,
could you please explain again how the process of "translating" the equation to b^2 - 4ac?
Thanks a lot
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discriminant

by pakaskwa » Wed Mar 04, 2009 4:56 am
Hey Baldini,

If quadratic formula ax^2 + bx + c = 0, its roots are:

x1,2 = [-b+/- square(b^2-4ac)] / 2

Only when (b^2-4ac) >= 0, there are real roots. Here, b^2-4ac is called discriminant.

For this question, a=1, b=2m+1, c=m^2-1. And you can do the rest.
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