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CAT 2019 Slot 1 — QA Question 26

Solving Quadratic EquationsEasy

The product of the distinct roots of ∣x2 − x − 6∣ = x + 2 is

Answer & solution

Correct answer: -16

  • -16

  • B

    -4

  • C

    -8

  • D

    -24

Solution

Easy

Factor inside the modulus: x2−x−6=(x−3)(x+2)x^2-x-6=(x-3)(x+2). The equation becomes ∣(x−3)(x+2)∣=x+2|(x-3)(x+2)|=x+2. Since the RHS must be ≥0\ge 0 we need x≥−2x\ge -2; split the modulus by the sign of (x−3)(x+2)(x-3)(x+2) on that range and collect all distinct roots.

1

Set up. Factor and note the domain restriction from the RHS.

∣x2−x−6∣=x+2 ⇒ ∣(x−3)(x+2)∣=x+2(factor)⇒ x+2≥0 ⇒ x≥−2(LHS≥0)\begin{aligned} &\left|x^2-x-6\right|=x+2\ \Rightarrow\ \left|(x-3)(x+2)\right|=x+2\quad\text{(factor)}\\ &\Rightarrow\ x+2\ge 0\ \Rightarrow\ x\ge -2\quad\text{(LHS}\ge 0) \end{aligned}
2

Case −2≤x≤3-2\le x\le 3. Here (x−3)(x+2)≤0(x-3)(x+2)\le 0, so the modulus flips sign.

−(x−3)(x+2)=x+2⇒ −(x−3)=1(divide by x+2, x≠−2)⇒ x=2(valid, lies in [−2,3])\begin{aligned} &-(x-3)(x+2)=x+2\\ &\Rightarrow\ -(x-3)=1\quad\text{(divide by }x+2,\ x\ne-2)\\ &\Rightarrow\ x=2\quad\text{(valid, lies in }[-2,3]) \end{aligned}

Also x+2=0x+2=0 gives the boundary root x=−2x=-2 (both sides 00), so x=−2x=-2 is a root.

3

Case x>3x>3. Here (x−3)(x+2)>0(x-3)(x+2)>0, modulus opens directly.

(x−3)(x+2)=x+2⇒ x−3=1(divide by x+2>0)⇒ x=4(valid, >3)\begin{aligned} &(x-3)(x+2)=x+2\\ &\Rightarrow\ x-3=1\quad\text{(divide by }x+2>0)\\ &\Rightarrow\ x=4\quad\text{(valid, }>3) \end{aligned}
4

Collect distinct roots and multiply. The roots are −2, 2, 4-2,\,2,\,4.

(−2)×2×4=−16\begin{aligned} &(-2)\times 2\times 4=-16 \end{aligned}
−16-16

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CAT 2019 Slot 1 QA Q26 — Solving Quadratic Equations solution | TheCATExam