Let m and n be the numbers of real roots of the quadratic equations x²-12 x+[x]+31=0 and x²-5|x+2|-4=0 respectively, where [x] denotes the greatest integer ≤ x. Then m²+mn+n² is equal to __________.

Solution

Maths sir · 4 steps

Pin down [x] before you solve anything, then check each root against the case it came from.

1. Remove the bracket. Write [x] = x - f, where f is the fractional part, 0 ≤ f < 1. The first equation becomes x² - 11x + 31 = f.

   0 ≤ x² - 11x + 31 < 1 ⇒ (x-5)(x-6) < 0 ⇒ 5 < x < 6

2. First equation. The left inequality always holds, because x² - 11x + 31 has discriminant 121 - 124 < 0. So any root lies in (5, 6), where [x] = 5. Then the equation is x² - 12x + 36 = 0, giving x = 6, which is outside (5, 6). So there is no real root, and m = 0.
3. Second equation. For x ≥ -2: x² - 5x - 14 = 0 gives x = 7 and x = -2, and both are allowed. For x < -2: x² + 5x + 6 = 0 gives x = -2 and x = -3, but only x = -3 is allowed. The distinct roots are 7, -2, -3, so n = 3.
4. So. m² + mn + n² = 0 + 0 + 9 = 9.

The common slip: 16. Counting x = -2 from both cases gives n = 4, and then the answer comes out as 16 instead of 9.

The idea to keep. With [x] or |x| in an equation, every candidate root must satisfy the condition of the case that produced it. Throw out the ones that do not, and count a shared root only once.

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