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Divide 20 into Two Parts Such that Three Times the Square of One Part Exceeds the Other Part by 10. - Mathematics

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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
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.

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APPEARS IN

 Selina Solution for Concise Mathematics for Class 10 ICSE (2020 (Latest))
Chapter 6: Solving (simple) Problems (Based on Quadratic Equations)
Exercise 6(A) | Q: 12 | Page no. 70
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Solution Divide 20 into Two Parts Such that Three Times the Square of One Part Exceeds the Other Part by 10. Concept: Quadratic Equations.
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