• Feb 17th 2011, 12:06 PM
BobBali
The question requires me to construct an equation and solve for y.

Q. When a mother's age is $y^2$ then the daughter's age is y. When the mother lives to the age of 13y then the daughter will be the same age of
$
y^2
$

Find the age of the daughter?

Ok here goes: $y^2 = y$
The mothers age >
$(y^2 + 13y)^2 = y^2$
$(y^2 + 13y) (y^2 + 13y) = y^2$
$y^4 +13y^3 + 13y^3 + 169y^2 = y^2$
$y^4 + 26y^3 + 168y^2 = 0$

$
y^2(y^2 + 26y + 168) = 0

$

Now solve using quadratic formula a=1, b=26, c=168

[tex] \frac{-26}{2} + - \frac{\sqrt{676-672}{2} [\MATH]

[tex] -13 + - \frac{\sqrt{4}{2} > x= -12 or x=-14 ?
[\MATH]

So, daughter's age is 12 or 14? But that doesn't make sense because if the daughter's age at the start was 12, then mother's age would be 144! Not feasible..(Thinking)
• Feb 17th 2011, 01:21 PM
emakarov
I don't understand how you came up with the equations.

Quote:

Originally Posted by BobBali
Ok here goes: $y^2 = y$

Where does this come from? Is this supposed to mean that y = 1?
Quote:

The mothers age >
$(y^2 + 13y)^2 = y^2$
Why do you add two mother's ages: the current one and the future one? And why do you square the result?

Mother's current age is $y^2$. After some number of years, she is $13y$.
Daugter's current age is $y$. After the same number of years, she is $y^2$.

From this, it seems clear how to construct an equation.
• Feb 17th 2011, 01:28 PM
Or...

the mother and daughter's age difference is constant

$y^2-y=13y-y^2$
• Feb 18th 2011, 11:00 AM
BobBali
Yes Sir, thank you! A hint was later given to us the next day that we should try and keep the ages of the mother and daughter constant; which apparently you already saw... Thank you once again. Take care.

Regards,
Bob
• Feb 18th 2011, 01:10 PM