Question

algebra 2 chapter
Example 1: Four friends have some coins. Pavan has 2 less than Samir, who has 2 less than Tarun. If Pranav has 2 less than Pavan, and the product of the number of coins that each of
them has is 5760, how many coins in all do they have among themselves?​

Answer 1

$Given \: Pavan , Samir, Tarun \:and \:Pranav$

$are \: four \: friends$

$Let \: number \: of \: coins \: Tarun \:have = x$

$number \: of \: coins \: Samir\:have = x-2$

$number \: of \: coins \: pavan\:have =( x-2-2)$

$= (x-4)$

$number \: of \: coins \: Pranav\:have =( x-4-2)$

$= (x-6)$

/* According to the problem given */

$Product \:of \: the \: coins = 5760$

$\implies x(x-2)(x-4)(x-6) = 5760$

$\implies x(x-6)(x-2)(x-4) = 5760$

$\implies (x^{2} - 6x )(x^{2} - 6x +8)-5760 = 0$

$Let \: x^{2} - 6x = a\:we \:have$

$a(a+8) - 5760 = 0$

$\implies a^{2} + 8a - 5760 = 0$

$\implies a^{2} - 72a + 80a - 5760 = 0$

$\implies a(a-72) + 80( a - 72 ) = 0$

$\implies (a-72)(a+80) = 0$

$\implies a - 72 = 0 \: a+80 = 0$

$\implies a = 72 \:Or \: a = -80$

$But a = x^{2} -6x = 72 \: --(1)$

$\implies x^{2} -12x +6x -72 = 0$

$\implies x(x-12)+6(x-12) = 0$

$\implies (x-12)(x+6) = 0$

$\implies x - 12 = 0 \: Or \: x + 6 = 0$

$\implies x = 12 \: Or \: x = -6$

/* x should not be negative */

$\green { x = 12 }$

$and \: a = x^{2} - 6x = -80$

$\implies x^{2} - 6x + 80 = 0$

$\implies\red{ Here \: we \:get \: imaginary \:roots}$

Therefore.,

$number \: of \: coins \: Tarun \:have = x = 12$

$number \: of \: coins \: Samir\:have = x-2$

$= 12 - 2 = 10$

$number \: of \: coins \: pavan\:have = = (x-4)$

$= 12-4 = 8$

$number \: of \: coins \: Pranav\:have =( x-6)$

$= 12-6 = 6$

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