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.
### (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.
