The age of a man is twice the square of the age of his son. 8 years hence the age of man will be 4 years more than three times the age of his son. Find their present ages
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see attachment...I hope it will help u...
the value of x comes to be +4 and - 5/2...so neglecting the negative value of x...becoz age can't be negative...so value of x is +4...
the value of x comes to be +4 and - 5/2...so neglecting the negative value of x...becoz age can't be negative...so value of x is +4...
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Answered by
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Let the age of man be x years
and age of his son be y years , then ;
x= 2×y^2
after 8 years,
the respective age of man will be = (x+8)yrs and age of his son will be (y+8)yrs , then;
(x+8)= 4+3(y+8)
2y^2 + 8 = 4 + 3y+24
2y^2 -3y -20 = 0
2y^2 -8y+5y -20 =0
2y(y-4) +5(y-4) = 0
(2y+5) (y-4) = 0
y = +4 or -5
since age can never be negative, therefore the age of the son will be 4 years.
Age of the man
= 2×y^2
= 2×16
=32 years
hope this helped you ..............:
and age of his son be y years , then ;
x= 2×y^2
after 8 years,
the respective age of man will be = (x+8)yrs and age of his son will be (y+8)yrs , then;
(x+8)= 4+3(y+8)
2y^2 + 8 = 4 + 3y+24
2y^2 -3y -20 = 0
2y^2 -8y+5y -20 =0
2y(y-4) +5(y-4) = 0
(2y+5) (y-4) = 0
y = +4 or -5
since age can never be negative, therefore the age of the son will be 4 years.
Age of the man
= 2×y^2
= 2×16
=32 years
hope this helped you ..............:
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