#### Question

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

#### Solution

Let the two parts be x and y.

From the given information,

x + y = 20 ⇒ y = 20 – x

3x^{2} = (20 – x) + 10

3x^{2} = 30 – x

3x^{2} + x – 30 = 0

3x^{2} – 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.

Is there an error in this question or solution?

Solution Divide 20 into Two Parts Such that Three Times the Square of One Part Exceeds the Other Part by 10. Concept: Quadratic Equations.