Question

One year ago age of father was 8 times the age of his child. Now the age of father is the square of the child's age . solve this using a second degree equation. Write every steps very clear. Find the present age of father and child??​

Answer 1

Answer:

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Step-by-step explanation:

$let \: the \: present \: age \: of \: father \: be \\ x \: years \: and \: that \: of \: his \: son \: be \: y \: years \\ \\ so \: according \: to \: first \: condition \\ x - 1 = 8(y - 1) \\ \\ x - 1 = 8y - 8 \\ \\ x = 8y - 7 \: \: \: \: \: \: \: (1) \\ \\ according \: to \: second \: condition \\ x = y {}^{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (2) \\ \\ so \: accordingly \\ from \: (1) \: and \: (2) \\ we \: get \\ \\ y {}^{2} = 8y - 7 \\ y {}^{2} - 8y + 7 = 0 \\ y {}^{2} - 7y - y + 7 = 0 \\ \\ y (y - 7) - 1(y - 7) = 0 \\ (y - 7)(y - 1) = 0 \\ \\ y = 7 \: \: or \: \: y = 1$

$but \: if \: we \: take \: y = 1 \\ then \\ \: x = y {}^{2} = (1) {}^{2} = 1 \\ \\ ie \: \: x = y \\ so \: as \: ages \: of \: father \: and \: son \: cant \: be \: equal \\ and \: moreover \: \: x > y \\ y = 1 \: \: is \: absurd \\ \\ hence \: then \\ y = 7 \\ \\ thus \: then \\ x = y {}^{2} \\ = (7) {}^{2} \\ x= 49$

$hence \: present \: age \: of \: father \: is \: 49 \: years \\ and \: his \: son \: is \: 7 \: years \: old$