Math, asked by vishalchouhanab4055, 1 year ago

Two years ago, a man's age was three times the square of his son's age. In three years time his age will be four times his son's age. Find their present ages

Answers

Answered by TeenTitansGo
11

Let the age of his son two years ago was x years and therefore according to the question his ( father's ) age would be 3 times the square of his son's age, so father's age = 3 x^2 years


Age of father 2 years ago = 3x^2  years

Age of his son 2 years ago = x years



Given that in three years age of father will be four times his son's age, according to the question 2 years their age was 3x^2 years and x years, so now their age will be 2 years more .

Present age of father = ( 3x^2 + 2 ) years

Present age of son = ( x + 2 ) years


Age of father after 3 years = ( 3x^2 + 2 + 3 ) years = ( 3x^2 + 5 ) years

Age of son after  3 years = ( x + 5 ) years


Given, Age of father after 3 years = 4 times the age of his son  

           

        (  3x^2 + 5 ) years = 4( x + 5 ) years

        3x^2 + 5  =  4x + 20

        3x^2 - 4x + 5 - 20 = 0

        3x^2 - 4x - 15 = 0

        3x^2 - ( 9 - 5 )x - 15 = 0

        3x^2 - 9x + 5x - 15 = 0

        3x( x - 3 ) + 5( x - 3 ) = 0

         ( x - 3 ) ( 3x + 5 ) = 0

 

So, x - 3 = 0  Or  3x + 5 = 0


x was assumed as age therefore age cannot be negative, value of x will be positive.

x = 3


Hence,

Present age of son = x years = 3 years

Present age of father = 3x^2 years = 3( 3 )^2 years = 27 years


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