(x2- 10 x) (x2- 49) <_0
To solve the inequality \((x^2 - 10x)(x^2 - 49) \leq 0\), we need to find the intervals where the product is less than or equal to zero.
Step 1: Factor the expression
First, we will factor both parts of the inequality:
- \(x^2 - 10x = x(x - 10)\)
- \(x^2 - 49 = (x - 7)(x + 7)\) (since it is a difference of squares)
So, we have:
\[ (x(x - 10))((x - 7)(x + 7)) \leq 0 \]
Step 2: Find the critical points
Next, we set each factor equal to zero to find the critical points:
- \(x = 0\)
- \(x - 10 = 0 \implies x = 10\)
- \(x - 7 = 0 \implies x = 7\)
- \(x + 7 = 0 \implies x = -7\)
The critical points are \(x = -7, 0, 7, 10\).
Step 3: Test intervals
Now we will test the sign of the product in the intervals defined by these critical points:
- \( (-\infty, -7) \)
- \( (-7, 0) \)
- \( (0, 7) \)
- \( (7, 10) \)
- \( (10, \infty) \)
We can pick a test point in each interval and calculate the sign of \((x^2 - 10x)(x^2 - 49)\):
For \(x = -8\) (in \((-∞, -7)\)): \[ (-8)((-8) - 10)((-8) - 7)((-8) + 7) \implies (-)(-)(-)(-) > 0 \]
For \(x = -1\) (in \((-7, 0)\)): \[ (-1)((-1) - 10)((-1) - 7)((-1) + 7) \implies (-)(-)(-)(+) < 0 \]
For \(x = 1\) (in \((0, 7)\)): \[ (1)((1) - 10)((1) - 7)((1) + 7) \implies (+)(-)(-)(+) > 0 \]
For \(x = 8\) (in \((7, 10)\)): \[ (8)((8) - 10)((8) - 7)((8) + 7) \implies (+)(-)(+)(+) < 0 \]
For \(x = 11\) (in \((10, ∞)\)): \[ (11)((11) - 10)((11) - 7)((11) + 7) \implies (+)(+)(+)(+) > 0 \]
Step 4: Summary of signs
From the signs we found:
- \( (-\infty, -7) : + \)
- \( (-7, 0) : - \)
- \( (0, 7) : + \)
- \( (7, 10) : - \)
- \( (10, \infty) : + \)
Step 5: Determine the solution
We are trying to find where the expression is less than or equal to zero:
- The intervals where the product is negative are \((-7, 0)\) and \((7, 10)\).
- We also need to include the points where the expression equals zero: \(x = -7, 0, 7, 10\).
Thus, the solution to the inequality is:
\[ \boxed{[-7, 0] \cup [7, 10]} \]
