b) The present age of the mother is equal to the square of the age of her daughter after one year. Likewise, the age of the daughter in 10 years will be 1 year less than the age of her mother 10 years ago.

i) Write the roots of the quadratic equation [tex]\( ax^2 + bx + c = 0, \, a \neq 0 \)[/tex].

ii) Find their present ages.

iii) How many years later will the mother's age exceed the daughter's age by 3 times the daughter's age plus 3 years?



Answer :

Sure, let us break down each part of the given problem step-by-step.

### (i) Write the roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex], [tex]\( a \neq 0 \)[/tex].
The roots of the general quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] can be found using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

### (ii) Find their present ages.
We have the following information:
1. The present age of the mother, [tex]\( M \)[/tex], is equal to the square of the age of her daughter, [tex]\( D \)[/tex], after one year:
[tex]\[ M = (D + 1)^2 \][/tex]
2. The age of the daughter in 10 years will be 1 year less than the age of her mother 10 years ago:
[tex]\[ D + 10 = M - 10 - 1 \][/tex]
[tex]\[ D + 10 = M - 11 \][/tex]

We need to solve these two equations simultaneously.

After solving, we find two sets of roots:
1. [tex]\( M = 16 \)[/tex] and [tex]\( D = -5 \)[/tex]
2. [tex]\( M = 25 \)[/tex] and [tex]\( D = 4 \)[/tex]

It is important to note that ages must be non-negative values. Out of the two sets of solutions, only the pair [tex]\( M = 25 \)[/tex] and [tex]\( D = 4 \)[/tex] makes sense.

Therefore, the present ages are:
[tex]\[ \text{Mother's age} = 25 \][/tex]
[tex]\[ \text{Daughter's age} = 4 \][/tex]

### (iii) How many years later, mother's age will exceed 3 years by 3 times the age of her daughter?
We need to find the number of years, [tex]\( y \)[/tex], after which the mother's age will be 3 years more than three times the daughter's age. Formulating this we get the equation:
[tex]\[ M + y = 3(D + y) + 3 \][/tex]

Plugging in the values of the present ages:
[tex]\[ 25 + y = 3(4 + y) + 3 \][/tex]
[tex]\[ 25 + y = 12 + 3y + 3 \][/tex]
[tex]\[ 25 + y = 15 + 3y \][/tex]

Solving for [tex]\( y \)[/tex]:
[tex]\[ 25 + y = 15 + 3y \][/tex]
[tex]\[ 25 - 15 = 3y - y \][/tex]
[tex]\[ 10 = 2y \][/tex]
[tex]\[ y = 5 \][/tex]

So, the mothers age will exceed the daughter’s age by 3 more than 3 times the daughters age after 5 years.