CAT 2023 Slot 1QA Question 15

Higher Degree PolynomialsEasy

The equation x3 + (2r + 1)x2 + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other roots are real, then the minimum possible non-negative integer value of r is?

Answer & solution

Answer: 2

Solution

Easy

Use Vieta's on the cubic. The product of roots forces the other two roots to be reciprocals; their sum is p+1pp+\tfrac1p, which is bounded away from (2,2)(-2,2). That inequality cages rr.

1

Name the other roots p,qp,q. Product of all roots of x3+(2r+1)x2+(4r1)x+2=0x^3+(2r+1)x^2+(4r-1)x+2=0 is 21-\dfrac{2}{1}:

(2)pq=2  pq=1  q=1p(-2)\cdot p\cdot q=-2\ \Rightarrow\ pq=1\ \Rightarrow\ q=\frac1p

So the three roots are 2, p, 1p-2,\ p,\ \dfrac1p.

2

Sum of roots =(2r+1)=-(2r+1):

2+p+1p=(2r+1) p+1p=2r+1\begin{aligned} &-2+p+\frac1p=-(2r+1)\\ &\Rightarrow\ p+\frac1p=-2r+1 \end{aligned}
3

For real pp, the quantity p+1pp+\dfrac1p never lies strictly between 2-2 and 22:

p+1p2orp+1p2p+\frac1p\ge 2\quad\text{or}\quad p+\frac1p\le -2

Hence 2r+12-2r+1\ge 2 or 2r+12-2r+1\le -2:

2r1  r122r3  r32\begin{aligned} &-2r\ge 1\ \Rightarrow\ r\le -\tfrac12\\ &-2r\le -3\ \Rightarrow\ r\ge \tfrac32 \end{aligned}
4

Least non-negative integer rr: the allowed band is r12r\le -\tfrac12 or r32r\ge \tfrac32. The smallest non-negative integer satisfying r32r\ge\tfrac32 is 22.

r=2r=\mathbf{2}
CAT 2023 Slot 1 QA Q15: The equation x 3 + (2r + 1)x 2 + (4r - 1)x + 2 = 0 has -2 as one of the roots. If the other roots are real, th — Solution | TheCATExam