Question

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

Answer 1

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

= x(3x + 10) – 3(3x + 10) = 0

= (3x + 10)(x – 3) = 0

= `x = 3, x = (-10)/3`

Since, x cannot be equal to `(-10)/3`, so, x = 3.

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