SOLUTION: The sum of the squares of Maria's two kids is 289. If the boy is seven years older than the girl, then what are their ages ?

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Question 190217This question is from textbook
: The sum of the squares of Maria's two kids is 289. If the boy is seven years older than the girl, then what are their ages ? This question is from textbook

Found 2 solutions by checkley75, nerdybill:
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
x^2+(x+7)^2=289
x^2+x^2+14x+49=289
2x^2+14x+49-289=0
2x^2+14x-240=0
2(x^2+7x-120)=0
2(x+15)(x-8)=0
x-8=0
x=8 ans. for the girl's age.
8+7=15 ans. for the boys age.
Proof:
8^2+15^2=289
64+225=289
289=289

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of the squares of Maria's two kids is 289. If the boy is seven years older than the girl, then what are their ages ?
.
Let g = age of the girl
then
g+7 = age of boy
.
g^2 + (g+7)^2 = 289
g^2 + (g+7)(g+7) = 289
g^2 + (g^2+14g+49) = 289
2g^2 + 14g + 49 = 289
2g^2 + 14g - 240 = 0
g^2 + 7g - 120 = 0
(g+15)(g-8) = 0
g = {8, -15}
We can throw away the negative solution leaving us with:
g = 8 years old (age of girl)
.
age of boy:
g+7 = 8+7 = 15 years old (age of boy)