Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 23 May 2013, 03:11

# For the equation x^2 + 2x + m = 5, where m is a constant, 3

Author Message
TAGS:
Intern
Joined: 01 Jun 2005
Posts: 19
Followers: 0

Kudos [?]: 0 [0], given: 0

For the equation x^2 + 2x + m = 5, where m is a constant, 3 [#permalink]  22 Jun 2005, 16:40
00:00

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
For the equation x^2 + 2x + m = 5, where m is a constant, 3 is one solution for x. What is the other solution?

a. -5
b. -2
c. -1
d. 3
e. 5

Can you explain the logic behind solving this problem? Thanks so much!
Intern
Joined: 22 Jun 2005
Posts: 11
Location: Austin, TX
Followers: 0

Kudos [?]: 0 [0], given: 0

a: -5

substitute the given solution of x, you can figure out that m = -10
you will get:
x^2 + 2x - 15 = 0
=> (x+5) (x-3) = 0
Similar topics Replies Last post
Similar
Topics:
For what values of m does the equation x^2-x+m=0 possess no 3 03 Mar 2004, 16:26
If M denotes the set of roots of equation 2-x^2 = (x-2)^2, 2 02 Dec 2007, 02:24
1 If 2x^2-3x-m is divisible by x-3, then m= -18 -9 -6 9 18 3 21 Mar 2008, 06:15
If 2x^2-3x-m is divisible by (x-3) what is m? 2 19 Jan 2012, 12:13
1 If set M consists of the root(s) of equation 2-x^2 = (x-2)^2 2 19 Nov 2012, 02:49
Display posts from previous: Sort by