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CAT Previous Year Paper | CAT Quant Questions | Question 7

CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 7 : The equation \(x^3+(2 r+1) x^2+(4 r-1) x+2=0\) has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of \(r\) is


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Explanatory Answer

\( x ^ { 3 } + ( 2 r + 1 ) x ^ { 2 } + ( 4 r - 1 ) x + 2 = 0 \)
Since -2 is one of the roots, the cubic equation can be factored as…
\( ( x + 2 ) \left( x ^ { 2 } + ( 2 x - 1 ) x + 1 \right) = 0 \)
Since the other two roots are real, \( \left( x ^ { 2 } + ( 2 x - 1 ) x + 1 \right) = 0 \) has two real roots.
That is the discriminant of \( \left( x ^ { 2 } + ( 2 x - 1 ) x + 1 \right) = 0 \) is non-negative.
\( ( 2 r - 1 ) ^ { 2 } \geq 4 \)
\( r \geq \frac { 3 } { 2 } \) or \( r \leq - \frac { 1 } { 2 } \)
Therefore the minimum possible non-negative integral value of r is 2.


The question is " The equation \(x^3+(2 r+1) x^2+(4 r-1) x+2=0\) has -2 as one of the roots. If the other two roots are real, then the minimum possible non-negative integer value of \(r\) is "

Hence, the answer is '2'

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