Started 3 weeks ago by Sumit Rawat in
7 Replies
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Replied 9 months ago
Hi, I have discussed this question in live doubt class earlier.I have attached the recording of that class .This is the 1st question in the recording.Link of the recording - https://dashboard.catking.in/courses/439/contents/196866/Let me know if you still have any doubt.Happy Learning!
Replied 3 weeks ago
Actual CAT 2023 SLOT 1 Question 7The equation x3+(2r+1)x2+(4r−1)x+2=0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r isCorrect Answer:2ExplanationAnswer- 2x3+(2r+1)x2+(4r−1)x+2=0Since -2 is one of the roots, the cubic equation can be factored as…(x+2)(x2+(2x−1)x+1)=0Since the other two roots are real, (x2+(2x−1)x+1)=0 has two real roots.That is the discriminant of (x2+(2x−1)x+1)=0 is non-negative.(2r−1)2≥4r≥32 or r≤−12Therefore the minimum possible non-negative integral value of r is 2.
Replied 3 weeks ago
Actual CAT 2023 SLOT 1 Question 7The equation x3+(2r+1)x2+(4r−1)x+2=0 has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of r isCorrect Answer:2ExplanationAnswer- 2x3+(2r+1)x2+(4r−1)x+2=0Since -2 is one of the roots, the cubic equation can be factored as…(x+2)(x2+(2x−1)x+1)=0Since the other two roots are real, (x2+(2x−1)x+1)=0 has two real roots.That is the discriminant of (x2+(2x−1)x+1)=0 is non-negative.(2r−1)2≥4r≥32 or r≤−12Therefore the minimum possible non-negative integral value of r is 2.