Question 331693
In three years, Janice will be three times as old as her daughter.
j + 3 = 3(d+3)
j + 3 = 3d + 9
j = 3d + 9 - 3
j = 3d + 6
:
 Six years ago, her age was her daughter's age squared.
j - 6 = (d-6)^2
j - 6 = d^2 - 12d + 36
Replace j with (3d+6)
3d + 6 - 6 = d^2 - 12d + 36
A quadratic equation
d^2 - 12d - 3d + 36 = 0
d^2 - 15d + 36 = 0
Factors to:
(d-12)(d-3) = 0
Two solutions
d = 3
d = 12 yrs is the solution that makes sense here
:
How old is Janice?
We know j = 3d + 6
j = 3(12) + 6
j = 42 yrs 
:
Check solutions in the equation: j - 6 = (d-6)^2
42 - 6 = (12-6)^2
36 = 6^2; confirms of our solutions
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