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Question 1209651: Determine the set of all real x satisfying
(x^2 - 5x + 3)^2 < 9 + 6x + x^2
Enter your answer in interval notation.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's how to solve the inequality:
1. **Rewrite the inequality:**
(x² - 5x + 3)² < (x + 3)²
2. **Take the square root of both sides:**
Since both sides are non-negative, we can take the square root:
|x² - 5x + 3| < |x + 3|
3. **Consider the different cases for the absolute values:**
Case 1: x² - 5x + 3 ≥ 0 and x + 3 ≥ 0 (i.e. x ≥ -3)
x² - 5x + 3 < x + 3
x² - 6x < 0
x(x - 6) < 0
0 < x < 6
Combining with x ≥ -3, we have 0 < x < 6.
Case 2: x² - 5x + 3 < 0 and x + 3 ≥ 0 (i.e. x ≥ -3)
-(x² - 5x + 3) < x + 3
-x² + 5x - 3 < x + 3
-x² + 4x - 6 < 0
x² - 4x + 6 > 0
The discriminant of x² - 4x + 6 is (-4)² - 4(1)(6) = 16 - 24 = -8 < 0. Since the leading coefficient is positive, the quadratic is always positive. So, this inequality is true for all real x.
Combining with x ≥ -3 and the condition that x² - 5x + 3 < 0, we need to find the roots of x² - 5x + 3 = 0.
x = (5 ± √(25 - 12))/2 = (5 ± √13)/2.
So, (5 - √13)/2 < x < (5 + √13)/2.
Since (5-√13)/2 ≈ 0.697 and (5+√13)/2 ≈ 4.303, and x ≥ -3, we have (5 - √13)/2 < x < (5 + √13)/2.
Case 3: x² - 5x + 3 ≥ 0 and x + 3 < 0 (i.e. x < -3)
x² - 5x + 3 < -(x + 3)
x² - 5x + 3 < -x - 3
x² - 4x + 6 > 0
As before, this inequality is true for all real x.
Combining with x < -3 and x² - 5x + 3 ≥ 0, we have x < (5 - √13)/2 and x < -3.
So, x < -3.
Case 4: x² - 5x + 3 < 0 and x + 3 < 0 (i.e. x < -3)
-(x² - 5x + 3) < -(x + 3)
-x² + 5x - 3 < -x - 3
-x² + 6x < 0
x² - 6x > 0
x(x - 6) > 0
x < 0 or x > 6.
Combining with x < -3 and the condition that (5 - √13)/2 < x < (5 + √13)/2, we have no solution in this case.
Combining all cases:
From Case 1: 0 < x < 6
From Case 2: (5 - √13)/2 < x < (5 + √13)/2
From Case 3: x < -3
Since (5-√13)/2 ≈ 0.697, the interval 0 < x < 6 covers the interval (5 - √13)/2 < x < (5 + √13)/2.
Final Answer: The final answer is $\boxed{(0, 6)}$
Answer by ikleyn(52903)  (Show Source):
You can put this solution on YOUR website! .
Determine the set of all real x satisfying
(x^2 - 5x + 3)^2 < 9 + 6x + x^2
Enter your answer in interval notation.
~~~~~~~~~~~~~~~~~~~~~~~~~~
Here I will solve this inequality by much simpler way
comparing with that in the post by @CPhill.
The starting inequality is
(x^2 - 5x + 3)^2 < 9 + 6x + x^2. (1)
Right side is (3+x)^2. So, we can rewrite the inequality in the form
(x^2 - 5x + 3)^2 < (3 + x)^2,
(x^2 - 5x + 3)^2 - (3 + x)^2 < 0.
Factor left side as the difference of squares. You will get
(x^2 - 5x + 3 - 3 - x)*(x^2 - 5x + 3 + 3 + x) < 0.
Combine like terms in parentheses
(x^2 - 6x)*(x^2 - 4x + 6) < 0. (2)
The quadratic trinomial x^2 - 4x + 6 has the discriminant d = b^2 - 4ac = (-4)^2 - 4*1*6 = 16 - 24 = -8.
Thus, the discriminant is negative - so, the quadratic trinomial x^2 - 4x + 6 is always positive
and, therefore, can be excluded from inequality (2).
Thus we should solve only
x^2 - 6x < 0,
or
x(x-6) < 0.
This inequality is true if and only if x and x-6 are of different signs.
It gives the ANSWER: 0 < x < 6, or, in the interval form, the solution set is this interval (0,6).
Solved.
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