Question

divide $20$ into two parts such that three time the square of one part exceeds the other pass by $10$

Answer 1

Given that, 20 divide into two parts such that three times the square of one part exceeds other by 10.

Let, the numbers are $x$ and $20-x$

Now,

$3x^2=(20-x)+10$

$3x^2=20-x+10$

$3x^2+x-30=0$

$3x^2+10x-9x-30=0$

$x(x+10)-3(x+10)=0$

$(x+10)(x-3)=0$

$x=-10or3$

-10 can’t be exist left it

Consider x=3

Now required numbers are x=3 and 20-x=17

Hence, this is the answer.