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MCQMediumJEE 2026Quadratic Equations in Complex Numbers

JEE Mathematics 2026 Question with Solution

The number of distinct real solutions of the equation x∣x+4∣+3∣x+2∣+10=0x|x + 4| + 3|x + 2| + 10 = 0 is

  • A

    22

  • B

    00

  • C

    33

  • D

    11

Answer

Correct answer:A

Step-by-step solution

Casewise solution using critical points

Given: The equation is x∣x+4∣+3∣x+2∣+10=0x|x + 4| + 3|x + 2| + 10 = 0.

Find: The number of distinct real solutions.

The equation involves absolute value expressions, so split the number line at the critical points x=−4x = -4 and x=−2x = -2.

Case I: x≥−2x \ge -2

Then ∣x+4∣=x+4|x + 4| = x + 4 and ∣x+2∣=x+2|x + 2| = x + 2.

Substituting,

x(x+4)+3(x+2)+10=0x(x + 4) + 3(x + 2) + 10 = 0 x2+4x+3x+6+10=0x^2 + 4x + 3x + 6 + 10 = 0 x2+7x+16=0x^2 + 7x + 16 = 0

Discriminant:

Δ=49−64=−15<0\Delta = 49 - 64 = -15 < 0

So, there is no real solution in this interval.

Case II: −4≤x<−2-4 \le x < -2

Then ∣x+4∣=x+4|x + 4| = x + 4 and ∣x+2∣=−(x+2)|x + 2| = -(x + 2).

Substituting,

x(x+4)+3(−x−2)+10=0x(x + 4) + 3(-x - 2) + 10 = 0 x2+4x−3x−6+10=0x^2 + 4x - 3x - 6 + 10 = 0 x2+x+4=0x^2 + x + 4 = 0

Discriminant:

Δ=1−16=−15<0\Delta = 1 - 16 = -15 < 0

So, there is no real solution in this interval.

Case III: x<−4x < -4

Then ∣x+4∣=−(x+4)|x + 4| = -(x + 4) and ∣x+2∣=−(x+2)|x + 2| = -(x + 2).

Substituting,

x(−x−4)+3(−x−2)+10=0x(-x - 4) + 3(-x - 2) + 10 = 0 −x2−4x−3x−6+10=0- x^2 - 4x - 3x - 6 + 10 = 0 −x2−7x+4=0-x^2 - 7x + 4 = 0

Multiplying by −1-1,

x2+7x−4=0x^2 + 7x - 4 = 0 x=−7±49+162=−7±652x = \frac{-7 \pm \sqrt{49 + 16}}{2} = \frac{-7 \pm \sqrt{65}}{2}

Both roots satisfy x<−4x < -4, so both are valid.

Therefore, there are exactly two distinct real solutions, so the correct option is A.

Common mistakes

  • A common mistake is not splitting the number line at x=−4x = -4 and x=−2x = -2. This is wrong because the signs of x+4x + 4 and x+2x + 2 change at these points. Instead, solve the equation separately on each interval.

  • Another mistake is using the wrong sign for the absolute value terms in the interval −4≤x<−2-4 \le x < -2. This is wrong because here ∣x+4∣=x+4|x + 4| = x + 4 but ∣x+2∣=−(x+2)|x + 2| = -(x + 2). Check the sign of each expression carefully before substituting.

  • Students may find roots of the quadratic in a case and count them without verifying the interval condition. This is wrong because roots obtained after removing absolute values are valid only if they lie in the interval assumed for that case. Always test the roots against the case restriction.

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