Question

One year ago a man was 8 times as old as his son, now his age is equal to the square of his sons age. Find the present age

Please solve this



If you don't know please don't answer ​

Answer 1

$\underline{\pink{ At \: present : }}$

$Age \: of \: son = x \: years$

$Age \: of \: father = x^{2} \: years$

$\underline{\pink{ One \: year \:ago : }}$

$Age \: of \: son = ( x - 1 )\: years$

$Age \: of \: father = (x^{2} - 1 ) \: years$

/* According to the problem given */

$x^{2} - 1 = 8(x-1)$

$\implies x^{2} - 1 - 8(x-1) = 0$

$\implies x^{2} - 1 - 8x + 1 = 0$

$\implies x^{2} - 8x = 0$

$\implies x(x-8) = 0$

$\implies x = 0 \:Or \: x = 8$

/* we should take x = 8 */

Therefore.,

$Present \:age \: son = 8 \: years$

$Present \:age \: father = 8^{2} = 64 \: years$

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