# CAT 2023 Question Paper | Quant Slot 1

###### 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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$x ^ { 3 } + ( 2 r + 1 ) x ^ { 2 } + ( 4 r - 1 ) x + 2 = 0$
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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