Question

find the real values of x which satisfy​

Answer 1

Answer:

Given, x²−3x+2>0 and x²−3x−4≤0

⇒(x−1)(x−2)>0 and (x−4)(x+1)≤0

⇒x∈(−∞,1)∪(2,∞) and x∈[−1,4]

Therefore, the common solution is [−1,1) ∪(2,4]

Answer 2

Answer:

x∈[-1,1)

Step-by-step explanation:

$=> x^{2} - 3x + 2 > 0\\ (x-2)(x-1) > 0$

.: x∈(-∞,1)U(2,∞);                         -(1)

$=> x^{2} - 3x - 4 \leq 0\\ (x+1)(x-4) \leq 0$

.: x∈[-1,4];                                    -(2)

From (1) and (2)

x∈[-1,1)U(2,4]

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