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

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

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22 Jun 2005, 16:40
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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!
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Joined: 22 Jun 2005
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22 Jun 2005, 17:16
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
22 Jun 2005, 17:16
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# For the equation x^2 + 2x + m = 5, where m is a constant, 3

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