The question requires me to construct an equation and solve for y.

Q. When a mother's age is $\displaystyle y^2$ 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

$\displaystyle

y^2

$

Find the age of the daughter?

Ok here goes: $\displaystyle y^2 = y$

The mothers age >

$\displaystyle (y^2 + 13y)^2 = y^2$

$\displaystyle (y^2 + 13y) (y^2 + 13y) = y^2$

$\displaystyle y^4 +13y^3 + 13y^3 + 169y^2 = y^2$

$\displaystyle y^4 + 26y^3 + 168y^2 = 0$

$\displaystyle

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..