Divide 20 into two parts such that three times the square of one part

1.	Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Answer» Let the two parts be x and y. From the given information, x + y = 20 ⟹y = 20 − x 3x² = (20 − x) + 10 3x² = 30 − x 3x² + x − 30 = 0 3x² − 9x + 10x − 30 = 0 3x(x − 3) + 10(x - 3) = 0 (x - 3) (3x + 10) = 0 x = 3, -10/3 Since, x cannot be equal to, -10/3 so, x=3 Thus, one part is 3 and other part is 20 − 3 = 17.

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

Answer»

Let the two parts be x and y.

From the given information,

x + y = 20 ⟹y = 20 − x

3x² = (20 − x) + 10

3x² = 30 − x

3x² + x − 30 = 0

3x² − 9x + 10x − 30 = 0

3x(x − 3) + 10(x - 3) = 0

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

x = 3, -10/3

Since, x cannot be equal to, -10/3 so, x=3

Thus, one part is 3 and other part is 20 − 3 = 17.