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CAT 2020 Slot 3 — QA Question 13

Discriminant and Roots of Quadratic EquationMedium

Let m and n be positive integers, If x² + mx + 2n = 0 and x² + 2nx + m = 0 have real roots, then the smallest possible value of m + n is

Answer & solution

Correct answer: 6

  • A

    8

  • B

    7

  • C

    5

  • 6

Solution

Medium

Each quadratic has real roots, so each discriminant is non-negative. This gives two inequalities linking mm and nn; combining them bounds nn from below, after which the least mm follows.

1

Discriminant of x2+mx+2n=0x^2+mx+2n=0.

m2−4(2n)≥0⇒ m2≥8n(real roots)(1)\begin{aligned} &m^2-4(2n)\ge 0\\ &\Rightarrow\ m^2\ge 8n \quad\text{(real roots)}\qquad(1) \end{aligned}
2

Discriminant of x2+2nx+m=0x^2+2nx+m=0.

(2n)2−4m≥0⇒ 4n2≥4m⇒ n2≥m(real roots)(2)\begin{aligned} &(2n)^2-4m\ge 0\\ &\Rightarrow\ 4n^2\ge 4m\\ &\Rightarrow\ n^2\ge m \quad\text{(real roots)}\qquad(2) \end{aligned}
3

Combine (1) and (2) to bound nn. From (2), m≤n2m\le n^2; from (1), m≥8nm\ge\sqrt{8n}.

n2≥m≥8n[(1) and (2)]⇒ n2≥8n⇒ n4≥8n(square both sides)⇒ n3≥8⇒ n≥2\begin{aligned} &n^2\ge m\ge \sqrt{8n} \quad\text{[(1) and (2)]}\\ &\Rightarrow\ n^2\ge \sqrt{8n}\\ &\Rightarrow\ n^4\ge 8n \quad\text{(square both sides)}\\ &\Rightarrow\ n^3\ge 8\\ &\Rightarrow\ n\ge 2 \end{aligned}
4

Least nn, then least mm. Take n=2n=2 and use (1).

m≥8×2=16=4⇒ mmin⁡=4(check (2): n2=4≥4=m ✓)\begin{aligned} &m\ge\sqrt{8\times 2}=\sqrt{16}=4\\ &\Rightarrow\ m_{\min}=4 \quad\text{(check (2): } n^2=4\ge 4=m\ \checkmark) \end{aligned}
m+n=4+2=6m+n=4+2=6

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CAT 2020 Slot 3 QA Q13 — Discriminant and Roots of Quadratic Equation solution | TheCATExam