Question

Construct the word problem on quadratic equation such that one of its answer is 10 (Years, natural numbers, Rs, etc).​

Answer 1

Answer:

your solution is as follows

pls mark it as brainliest

Step-by-step explanation:

$the \: sum \: of \: present \: ages \: of \: mother \\ and \: that \: of \: her \: son \: is \: 40. \\ three \: years \: ago \: the \: product \: of \: their \\ ages \: was \: 189. \\ find \: their \: respective \: present \: ages \\ \\ solution : \\ let \: the \: present \: age \: of \: mother \: be \: x \: years \\ and \: that \: of \: her \: son \: be \: y \: years \\ \\ according \: to \: first \: condition \\ x + y = 40 \\ x = 40 - y \: \: \: \: \: \: (1) \\ \\ according \: to \: first \: condition \\ (x - 3)(y - 3) = 189 \\ xy - 3x - 3y + 9 = 189 \\ \\ y(40 - y) - 3(40 - y) - 3y = 180 \\ 40y - y {}^{2} - 120 + 3y - 3y = 180 \\ \\ 40y - y { }^{2} = 300 \\ y {}^{2} - 40y + 300 = 0 \\ \\ y {}^{2} - 30y -10 y + 300 = 0 \\ y(y - 30) - 10(y - 30) = 0 \\ (y - 30)(y - 10) = 0 \\ \\ y = 30 \: \: or \: \: y = 10$

$but \: if \: we \: take \: \\ y = 30 \\ then \: we \: have \\ \: x = 40 - y = 40 - 30 = 10 \\ which \: is \: age \: of \: mother \\ \\ so \: since \: \\ age \: of \: mother > age \: of \: her \: son \\ y = 30 \: \: is \: absurd \\ \\ thus \: then \\ y = 10 \\ hence \: then \\ x = 30$