The sum of the squares of two consecutive odd positive integers is 290.
With smaller integer $x$ , the equation simplifies to $x^{2} + 2 x - 143 = 0$ .
Find the smaller integer.

Question

The sum of the squares of two consecutive odd positive integers is 290. With smaller integer x, the equation simplifies to x^2 + 2x - 143 = 0. Find the smaller integer.

Accepted Answer

11. x^2 + (x+2)^2 = 290 gives 2x^2 + 4x - 286 = 0, i.e. x^2 + 2x - 143 = 0. x = -2 ± √4 + 5722 = -2 ± 24/2 = 11. Integers 11 and 13.

The sum of the squares of two consecutive odd positive integers is 290.
With smaller integer $x$ , the equation simplifies to $x^{2} + 2 x - 143 = 0$ .
Find the smaller integer.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 The Quadratic Formula practice

The sum of the squares of two consecutive odd positive integers is 290.With smaller integer x, the equation simplifies to x2+2x−143=0.Find the smaller integer.

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 The Quadratic Formula practice

The sum of the squares of two consecutive odd positive integers is 290.
With smaller integer $x$ , the equation simplifies to $x^{2} + 2 x - 143 = 0$ .
Find the smaller integer.