Question

divide 20 into two parts such that three times the square of one part exceeds the other part by10

Answer 1

Heya user
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

verified answer

Step-by-step explanation:

Let one part be X.

So,the other part is (20-x)

ATQ,

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

3x^2=30-x

3x^2+10x-30=0

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

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

(3x+10)(×-3)=0

So,x=-10/3 or 3

So, two parts are 17 and 3

.open it helps!!