Consider the equation x²+4 x-n=0, where n ∈[20,100] is a natural number. Then the number of all distinct values of n, for which the given equation has integral roots, is equal to

Solution

Maths sir · 3 steps

Complete the square: the roots are -2 ± √(n+4), so make n+4 a perfect square.

1. Complete the square. Rewrite x² + 4x - n = 0 as (x+2)² = n + 4. This form gives the roots directly.

   x = -2 ± √(n+4)

2. Condition for integers. The roots are integers exactly when √(n+4) is an integer, that is, when n+4 is a perfect square. Since 20 ≤ n ≤ 100, you need a perfect square with 24 ≤ n+4 ≤ 104.

   n+4	n	Roots -2 ± √(n+4)
   25 = 5²	21	3, -7
   36 = 6²	32	4, -8
   49 = 7²	45	5, -9
   64 = 8²	60	6, -10
   81 = 9²	77	7, -11
   100 = 10²	96	8, -12

3. So. The squares from 5² to 10² all fit, which gives 6 values of n. The answer is (a).

The idea to keep. For a monic quadratic with integer coefficients, integral roots need a perfect-square discriminant. Completing the square shows the exact square you need, so you only have to count squares in a range.

Answer from the NTA final answer key. Solution by PrepWiser.