TS EAMCET · Maths · Quadratic Equation
The set of solutions satisfying both \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4 < 0\) is
- A \((-4,1)\)
- B \((-4,-3] \cup[-2,1)\)
- C \((-4,-3) \cup(-2,1)\)
- D \([-4,-3] \cup[-2,1]\)
Answer & Solution
Correct Answer
(B) \((-4,-3] \cup[-2,1)\)
Step-by-step Solution
Detailed explanation
Given, \(x^2+5 x+6 \geq 0\) and \(x^2+3 x-4 < 0\) \[ \begin{aligned} & \Rightarrow x \in(-\infty,-3] \cup[-2, \infty) \text { and } x \in(-4,1) \\ & \end{aligned} \] Common condition is \[ x \in(-4,-3] \cup[-2,1) \]
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