X^2-4x+6>|x+2|

Question 1

Задача Задача №2213098

А

Аноним

13 ноября 2024 03:48 · публичный вопрос

А

Аноним 13 ноября 2024 03:48

X^2-4x+6>|x+2|

Answer

To solve the inequality \(x^2 - 4x + 6 > |x + 2|\), we need to consider the expression for \(|x + 2|\) in two cases: when \(x + 2 \geq 0\) (which means \(x \geq -2\)), and when \(x + 2 < 0\) (which means \(x < -2\)).

Case 1: \(x + 2 \geq 0\) (i.e. \(x \geq -2\))

In this case, \(|x + 2| = x + 2\). The inequality becomes: \[ x^2 - 4x + 6 > x + 2 \] Rearranging gives: \[ x^2 - 4x - x + 6 - 2 > 0 \] \[ x^2 - 5x + 4 > 0 \] Factoring the quadratic: \[ (x - 1)(x - 4) > 0 \] To find the critical points, set each factor to zero:

\(x - 1 = 0 \implies x = 1\)
\(x - 4 = 0 \implies x = 4\)

Now, we need to test intervals based on these points: \((- \infty, 1)\), \((1, 4)\), and \((4, +\infty)\).

For \(x < 1\), say \(x = 0\): \((0 - 1)(0 - 4) = 1 > 0\) (True).
For \(1 < x < 4\), say \(x = 2\): \((2 - 1)(2 - 4) = (-2) < 0\) (False).
For \(x > 4\), say \(x = 5\): \((5 - 1)(5 - 4) = 4 > 0\) (True).

Thus, in this case, the solution is \(x \in (-\infty, 1) \cup (4, +\infty)\).

Case 2: \(x + 2 < 0\) (i.e., \(x < -2\))

In this case, \(|x + 2| = -(x + 2) = -x - 2\). The inequality becomes: \[ x^2 - 4x + 6 > -x - 2 \] Rearranging gives: \[ x^2 - 4x + x + 6 + 2 > 0 \] \[ x^2 - 3x + 8 > 0 \] To find the roots of the quadratic \(x^2 - 3x + 8\), we use the discriminant: \[ D = (-3)^2 - 4 \cdot 1 \cdot 8 = 9 - 32 = -23 \] Since the discriminant is negative, \(x^2 - 3x + 8\) has no real roots and is always positive (since the coefficient of \(x^2\) is positive).

Thus, for \(x < -2\), the inequality holds for all \(x\) in that range. Therefore, in this case, the solution is \(x < -2\).

Final Solution

Combining the results from both cases:

From Case 1: \(x \in (-\infty, 1) \cup (4, +\infty)\).
From Case 2: \(x < -2\).

The overall solution to the inequality \(x^2 - 4x + 6 > |x + 2|\) is: \[ x \in (-\infty, -2) \cup (4, +\infty). \]

Question 2

x^2-4x+6>|x+2|

Answer

To solve the inequality \(x^2 - 4x + 6 > |x + 2|\), we need to consider the expression for \(|x + 2|\) in two cases: when \(x + 2 \geq 0\) (which means \(x \geq -2\)), and when \(x + 2 < 0\) (which means \(x < -2\)). ### Case 1: \(x + 2 \geq 0\) (i.e. \(x \geq -2\)) In this case, \(|x + 2| = x + 2\). The inequality becomes: \[ x^2 - 4x + 6 > x + 2 \] Rearranging gives: \[ x^2 - 4x - x + 6 - 2 > 0 \] \[ x^2 - 5x + 4 > 0 \] Factoring the quadratic: \[ (x - 1)(x - 4) > 0 \] To find the critical points, set each factor to zero: - \(x - 1 = 0 \implies x = 1\) - \(x - 4 = 0 \implies x = 4\) Now, we need to test intervals based on these points: \((- \infty, 1)\), \((1, 4)\), and \((4, +\infty)\). - For \(x < 1\), say \(x = 0\): \((0 - 1)(0 - 4) = 1 > 0\) (True). - For \(1 < x < 4\), say \(x = 2\): \((2 - 1)(2 - 4) = (-2) < 0\) (False). - For \(x > 4\), say \(x = 5\): \((5 - 1)(5 - 4) = 4 > 0\) (True). Thus, in this case, the solution is \(x \in (-\infty, 1) \cup (4, +\infty)\). ### Case 2: \(x + 2 < 0\) (i.e., \(x < -2\)) In this case, \(|x + 2| = -(x + 2) = -x - 2\). The inequality becomes: \[ x^2 - 4x + 6 > -x - 2 \] Rearranging gives: \[ x^2 - 4x + x + 6 + 2 > 0 \] \[ x^2 - 3x + 8 > 0 \] To find the roots of the quadratic \(x^2 - 3x + 8\), we use the discriminant: \[ D = (-3)^2 - 4 \cdot 1 \cdot 8 = 9 - 32 = -23 \] Since the discriminant is negative, \(x^2 - 3x + 8\) has no real roots and is always positive (since the coefficient of \(x^2\) is positive). Thus, for \(x < -2\), the inequality holds for all \(x\) in that range. Therefore, in this case, the solution is \(x < -2\). ### Final Solution Combining the results from both cases: - From Case 1: \(x \in (-\infty, 1) \cup (4, +\infty)\). - From Case 2: \(x < -2\). The overall solution to the inequality \(x^2 - 4x + 6 > |x + 2|\) is: \[ x \in (-\infty, -2) \cup (4, +\infty). \]