Question

1. The present ages of a father and his on are 40 years and 8 years resepectively.
How many years ago, the product of their ages was 105? Find it.​

Answer 1

Answer:

let the number of years ago = x

(40-x)(8-x)=105

$320-48x+x^2=105$

$320-105-48x+x^2=0$

$x^2-48x+215=0$

$x^2-43x-5x+215=0$

$x(x-43)-5(x-43) = 0$

$(x-5)(x-43)=0$

$x=5,x=43$

but x cannot be 43 as father is currently 40

therefore,

     5 many years ago, the product of their ages was 105.

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