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Basic Algebra question

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by ctgambler » Sat Jul 11, 2009 5:09 pm
I'm a little stuck on the mechanics of a question - and it seems basic. Anyway, here it is.

In three years, Janice will be three times as old as her daughter. Six years ago, her age was her daughter's age squared. How old is Janice?

If someone could please show me the steps...I'd so appreicate it. thanks
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by truplayer256 » Sat Jul 11, 2009 5:30 pm
J= Janice's age
D= Daughter's age

J+3=3(D+3)

J-6=(D-6)^(2)

3D+6-6= (D-6)^(2)

3D=D^2-12D+36

D^2-15D+36=0

15+/-sqrt(225-144)/2=15+/-9/2

15+9/2=12

15-9/2=3

D=3 or D=12

Janice could be:

J+3=3(6)

J=15 years old

-or-

J+3=3(15)

J=42 yrs old.
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