Question 331693: Dear Sir/Madam; Pls. kindly see the problem below:
Question: 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?
I wrote out an Algebraic expression for this problem as follows: denoting the daughter's age by x to be able to solve for her daughter's age first, so Janice's age in three years would be: 3x+3, and her age six years ago would have been x2-6; x2-6+3x+3=0, which when simplified will look like this; x2+3x-3=0. But I do not think I am right. Could you please help. Thank you. Found 2 solutions by email@example.com, mananth:Answer by firstname.lastname@example.org(15639) (Show Source):
You can put this solution on YOUR website! 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
(d-12)(d-3) = 0
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
You can put this solution on YOUR website! let daughter's age be x now.
after 3 years daughter will be x+3
after 3 years Janice will be 3*(x+3) years
6 years ago daughter's age = x-6.
janice ' age was (x-6)^2
x=12 OR 3
After 3 years Janice will be x(x+3)
plug value of x
So now Janice will be 45-3 = 42 years ( this makes sense)
For value x=3
plug value of 3
3(3+3)= 18 This was 3 years later
now she will be 15