Question

If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now?​

Answer 1

let the actual age of Zeba = x years.
Her age when she was 5 years younger $=(x-5)$ years.
According to the given problem,
Square of her age = 11 years more than five times her actual age.
$\Rightarrow (x-5{)}^2=5\times \text{actual age}+11$
$\Rightarrow (x-5{)}^2=5x+11$
$|\because (a-b{)}^2=a^2+b^2-2ab|$
$\Rightarrow x^2-15x+14=0$
On splitting the middle term, we get
$x^2-14x-x+14=0$
$\Rightarrow x(x-14)-1(x-14)=0$
$\Rightarrow (x-14)(x-1)=0$
$\Rightarrow x=14.$
Here, x $\ne$1, because her age is $x-5$. so, $x-5=1-5=-4$ i.e. age cannot be negative
Hence, present age of Zeba is 14 years.