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CAT 2019 Slot 1 — QA Question 15

Solving Quadratic EquationsEasy

The number of solutions to the equation |x|(6x2 + 1) = 5x2 is

Answer & solution

Correct answer: 5

Answer: 5

Solution

Easy

The equation is ∣x∣(6x2+1)=5x2|x|(6x^2+1)=5x^2. Because of ∣x∣|x|, split into x>0x>0, x=0x=0, x<0x<0. In each non-zero case divide by xx (or ∣x∣|x|) to get a quadratic in xx, count its roots, and add up. Symmetry between the two cases makes the count quick.

1

Case x>0x>0. Here ∣x∣=x|x|=x; divide by xx.

x(6x2+1)=5x2⇒ 6x2−5x+1=0(divide by x)⇒ x=12, 13(both positive — valid)\begin{aligned} &x(6x^2+1)=5x^2\\ &\Rightarrow\ 6x^2-5x+1=0\quad\text{(divide by }x\text{)}\\ &\Rightarrow\ x=\tfrac{1}{2},\ \tfrac{1}{3}\quad\text{(both positive — valid)} \end{aligned}
2

Case x=0x=0. Check directly.

LHS=0⋅1=0,RHS=0⇒ x=0 is a solution\begin{aligned} &\text{LHS}=0\cdot1=0,\quad \text{RHS}=0\\ &\Rightarrow\ x=0 \text{ is a solution} \end{aligned}
3

Case x<0x<0. Here ∣x∣=−x|x|=-x; divide by xx.

−x(6x2+1)=5x2⇒ 6x2+5x+1=0(divide by x)⇒ x=−12, −13(both negative — valid)\begin{aligned} &-x(6x^2+1)=5x^2\\ &\Rightarrow\ 6x^2+5x+1=0\quad\text{(divide by }x\text{)}\\ &\Rightarrow\ x=-\tfrac{1}{2},\ -\tfrac{1}{3}\quad\text{(both negative — valid)} \end{aligned}
4

Total count. Add the valid solutions from all three cases.

2 (from x>0)+1 (x=0)+2 (from x<0)=5\begin{aligned} &2\ (\text{from }x>0)+1\ (x=0)+2\ (\text{from }x<0)=5 \end{aligned}
Number of solutions=5\text{Number of solutions}=5

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CAT 2019 Slot 1 QA Q15: The number of solutions to the equation |x|(6 x 2 + 1) = 5 x 2 is — Solution | TheCATExam