Math  /  Algebra

Question1x+2+3x2\frac{1}{x+2}+\frac{3}{x} \leqslant-2

Studdy Solution

STEP 1

1. The domain of x is all real numbers except for x = 0 and x = -2, as these values would make the denominators zero.
2. We are dealing with an inequality, so the direction of the inequality may change when we multiply or divide by a negative number.
3. We will need to find a common denominator to combine the fractions on the left side of the inequality.

STEP 2

1. Find a common denominator and combine the fractions on the left side.
2. Multiply both sides by the common denominator to eliminate fractions.
3. Rearrange the inequality to standard form.
4. Solve the quadratic inequality.

STEP 3

To combine the fractions, we need to find a common denominator. The common denominator will be (x+2)(x):
1x+2+3x=xx(x+2)+3(x+2)x(x+2) \frac{1}{x+2} + \frac{3}{x} = \frac{x}{x(x+2)} + \frac{3(x+2)}{x(x+2)}
Now we can add the numerators:
x+3x+6x(x+2)=4x+6x(x+2) \frac{x + 3x + 6}{x(x+2)} = \frac{4x + 6}{x(x+2)}
Our inequality now looks like:
4x+6x(x+2)2 \frac{4x + 6}{x(x+2)} \leq -2

STEP 4

Multiply both sides of the inequality by x(x+2). Note that we don't know the sign of x(x+2), so we need to consider two cases:
Case 1: If x(x+2) > 0, the inequality direction remains the same: 4x+62x(x+2) 4x + 6 \leq -2x(x+2)
Case 2: If x(x+2) < 0, the inequality direction changes: 4x+62x(x+2) 4x + 6 \geq -2x(x+2)

STEP 5

Let's rearrange the inequality to standard form for both cases:
Case 1: x(x+2) > 0 4x+62x24x 4x + 6 \leq -2x^2 - 4x 2x2+8x+60 2x^2 + 8x + 6 \leq 0
Case 2: x(x+2) < 0 4x+62x24x 4x + 6 \geq -2x^2 - 4x 2x2+8x+60 2x^2 + 8x + 6 \geq 0

STEP 6

Now we need to solve these quadratic inequalities:
For Case 1: 2x^2 + 8x + 6 ≤ 0 The quadratic formula gives us roots at x = -3 and x = -1. The parabola opens upward (coefficient of x^2 is positive). So, the solution for this case is: -3 ≤ x ≤ -1
For Case 2: 2x^2 + 8x + 6 ≥ 0 This is the opposite of Case 1, so the solution is: x ≤ -3 or x ≥ -1
Now, we need to combine these results with the conditions from Step 2:
Case 1 is valid when x(x+2) > 0, which occurs when x < -2 or x > 0 Case 2 is valid when x(x+2) < 0, which occurs when -2 < x < 0
Combining all of this information:
For x < -2: We use Case 2, so x ≤ -3 is the solution. For -2 < x < 0: We use Case 2, but there are no solutions in this interval. For x > 0: We use Case 1, but there are no solutions in this interval.
Therefore, the final solution is:
x3 x \leq -3
The solution to the inequality is:
x3 \boxed{x \leq -3}

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