SOLUTION: a mother has two children whose ages differ by 5 years. the sum of the squares of their ages is 97. the square of the mothers age can be found by writing the squares of the childre

Question 1015523: a mother has two children whose ages differ by 5 years. the sum of the squares of their ages is 97. the square of the mothers age can be found by writing the squares of the childrens ages one after the other as a four digit number. how old is the mother?
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Let the children's age be a & b
Let m = mom's age
:
write an equation for each statement
:
a mother has two children whose ages differ by 5 years.
a - b = 5
a = (b+5)
the sum of the squares of their ages is 97.
a^2 + b^2 = 97
Replace a with (b+5)
(b+5)^2 + b = 97
FOIL (b+5)(b+5)
b^2 + 10b + 25 + b^2 = 97
Combine like terms
2b^2 + 10b + 25 - 97 = 0
2b^2 + 10b - 72 = 0
Factors to
(2b + 18)(b - 4) = 0
The positive solution
b = 4 yrs
then
a = 4 + 5 = 9 yrs
Check it by finding the sum of the squares of their ages
4^2 + 9^2 =
16 + 81 = 97
:
the square of the mothers age can be found by writing the squares of the children's ages one after the other as a four digit number.
Mom has to be less than 60 that would be 3600 when it is squared,
it couldn't be 8116, therefore:
m =
m = 41 is mom's age

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