SOLUTION: Determine the sum of all real numbers x satisfying (x^2 - 6x + 4)(x^2

SOLUTION: Determine the sum of all real numbers x satisfying (x^2 - 6x + 4)(x^2 - 8x + 5) = 1.

Question 1209657: Determine the sum of all real numbers x satisfying
(x^2 - 6x + 4)(x^2 - 8x + 5) = 1.
Answer by proyaop(69) (Show Source): You can put this solution on YOUR website!
Here's how to find the sum of all real numbers x satisfying the given equation:
1. **Rewrite the equation:**
(x² - 6x + 4)(x² - 8x + 5) = 1
2. **Substitution (optional):**
While not strictly necessary, you could substitute u = x² - 7x to try and simplify. However, it's easier to proceed without it.
3. **Expand and rearrange:**
Expanding the left side gives:
x⁴ - 8x³ + 5x² - 6x³ + 48x² - 30x + 4x² - 32x + 20 = 1
x⁴ - 14x³ + 57x² - 62x + 19 = 0
4. **Vieta's Formulas:**
For a quartic equation of the form ax⁴ + bx³ + cx² + dx + e = 0, the sum of the roots is given by -b/a. In our case, a = 1 and b = -14.
Therefore, the sum of all roots (real and complex) is -(-14)/1 = 14.
5. **Consider the possibility of complex roots:**
A quartic equation can have up to four roots. Some of these roots could be complex. However, Vieta's formulas tell us the sum of *all* roots is 14.
6. **Realizing that all roots must be real:**
Let f(x) = (x^2 - 6x + 4)(x^2 - 8x + 5) - 1.
Note that f(1) = (1-6+4)(1-8+5) - 1 = (-1)(-2) - 1 = 1 > 0.
f(2) = (4 - 12 + 4)(4 - 16 + 5) - 1 = (-4)(-7) - 1 = 27 > 0.
f(3) = (9 - 18 + 4)(9 - 24 + 5) - 1 = (-5)(-10) - 1 = 49 > 0.
f(4) = (16 - 24 + 4)(16 - 32 + 5) - 1 = (-4)(-11) - 1 = 43 > 0.
f(5) = (25 - 30 + 4)(25 - 40 + 5) - 1 = (-1)(-10) - 1 = 9 > 0.
f(6) = (36 - 36 + 4)(36 - 48 + 5) - 1 = 4(-7) - 1 = -29 < 0.
f(7) = (49 - 42 + 4)(49 - 56 + 5) - 1 = 11(-2) - 1 = -23 < 0.
f(8) = (64 - 48 + 4)(64 - 64 + 5) - 1 = 20(5) - 1 = 99 > 0.
Since f(x) is a continuous function, there must be a root between 6 and 8.
Also f(0) = 19 > 0.
Since the leading term is x^4, it goes to infinity as x goes to infinity and negative infinity.
Since f(1) > 0, f(6) < 0, f(8) > 0, there are two real roots between 1 and 8.
Since f(0) > 0 and f(x) goes to infinity as x goes to negative infinity, there must be a real root less than 0.
Since the sum of roots is 14, and there are 4 roots, all must be real.
7. **Final Answer:**
The sum of all real roots is 14.

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