Question

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

Answer 1

Hey

Here is your answer,

Let one part be x .
So , the other part is (20 -x)

ATQ ,
3x^2 = (20 - x) + 10
=> 3x^2 = 30 - x
=> 3x^2 + x - 30 = 0
=> 3x^2 + 10x - 9x - 30 = 0
=> 3x ( x - 3 ) + 10 ( x - 3 ) = 0
=> ( 3x + 10 ) ( x - 3 ) = 0

So , x = -10/3 or 3 .

Since , part of anything cant be (-)ve , x = 3 .

So , the two parts are 17 and 3 .

Hope it helps !!

Answer 2

Let the two parts are x & y.
So, according to the question 3 times the square of one part which is
$3x {}^{2}$
And another one is y.
So,
$3x {}^{2} > y$
Hence,
$3x {}^{2} - y =10 \\ = > 3x {}^{2} - y - 10 = 0$
And now put y=(20-x)
Hope you get the answers.