mary is now four times as old as her daugter.Four years ago the product of their ages was 40.Find present ages.
Daughter is X yrs. old.
Mary is 4x yrs. old.
(x-4)*(4x-4) = 40.
4x^2-4x-16x+16 = 40,
4x^2-20x-24 = 0,
x^2-5x-6 = 0, -6 = -6*1. sum = -5 = B.
(x-6)(x+1) = 0.
x-6 = 0. X = 6.
x+1 = 0. x = -1.
Use X = 6.
4x = 4*6 = 24.
m = 4d
(m-4)(d-4) = 40
Now just solve for m and d.
Final answer is....
To solve this problem, let's assign some variables:
Let M be the age of Mary.
Let D be the age of her daughter.
We are given two pieces of information:
1. "Mary is now four times as old as her daughter." translates to the equation: M = 4D.
2. "Four years ago the product of their ages was 40." translates to (M-4) * (D-4) = 40.
Now, let's solve the equations:
From equation 1, we have M = 4D.
Substituting this value into equation 2, we get:
(4D - 4) * (D - 4) = 40
Expanding and simplifying:
4D^2 - 16D - 4D + 16 = 40
4D^2 - 20D + 16 = 40
Rearranging:
4D^2 - 20D - 24 = 0
Dividing through by 4:
D^2 - 5D - 6 = 0
Factoring the quadratic equation:
(D - 6)(D + 1) = 0
So, D = 6 or D = -1.
Since we are dealing with ages, the daughter's age cannot be negative. Therefore, D = 6.
Using equation 1 to find M:
M = 4D = 4 * 6 = 24.
Thus, the present ages are Mary = 24 and her daughter = 6.