The number of distinct real roots of the equation |x+1||x+3|-4|x+2|+5=0, is _______

Solution

Maths sir · 4 steps

Everything sits around x = -2, so put t = |x+2| and the equation turns into one in t.

1. Spot the centre. Note that (x+1)(x+3) = (x+2)² - 1. Put t = |x+2| ≥ 0. Then |x+1||x+3| = |t² - 1|, and the equation becomes |t² - 1| - 4t + 5 = 0.
2. First case. For t ≥ 1, |t² - 1| = t² - 1, so t² - 4t + 4 = 0, that is (t-2)² = 0. This gives t = 2, which satisfies t ≥ 1, so we keep it.
3. Second case. For 0 ≤ t < 1, |t² - 1| = 1 - t², so t² + 4t - 6 = 0 and t = -2 ± √10. The positive value is -2 + √10 ≈ 1.16, which is not less than 1, so we reject it. The other value is negative, so it is rejected too.

   |x+2| = 2 ⇒ x = 0 or x = -4

4. So. Check: at x = 0, 1 × 3 - 8 + 5 = 0, and at x = -4, 3 × 1 - 8 + 5 = 0. The equation has 2 distinct real roots.

The common slip: 4. Keeping t = -2 + √10 without checking that it lies in its own case (t < 1) adds two false values of x and gives 4.

The idea to keep. When the factors are x+a and x+b, shift to the midpoint c = (a+b)/2: their product is (x+c)² minus a constant. Then always check each root against the range of its own case.

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