Solve` 8x^4 -2x^3 - 27 x^2 +6x +9=0` given that two roots have the same absolute value , but are opposite in sign.

Answer

To solve the equation \( 8x^4 - 2x^3 - 27x^2 + 6x + 9 = 0 \) given that two roots have the same absolute value but are opposite in sign, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \( a, -a, b, c \). Here, \( a \) and \( -a \) are the two roots with the same absolute value but opposite signs.

### Step 2: Use Vieta's Formulas
From Vieta's formulas, we know:
1. The sum of the roots \( S_1 = a + (-a) + b + c = 0 \).
2. The sum of the products of the roots taken two at a time \( S_2 = ab + ac + (-a)b + (-a)c = -\frac{-2}{8} = \frac{1}{4} \).
3. The sum of the products of the roots taken three at a time \( S_3 = -a^2b - a^2c + abc = -\frac{6}{8} = -\frac{3}{4} \).
4. The product of the roots \( S_4 = -a^2bc = \frac{9}{8} \).

### Step 3: Simplify the Equations
From \( S_1 \):
\[ 0 = 0 \quad \text{(which is always true)} \]

From \( S_2 \):
\[ ab + ac - ab - ac = 0 \Rightarrow b + c = \frac{1}{4} \quad \text{(Equation 1)} \]

From \( S_3 \):
\[ -a^2b - a^2c + abc = -\frac{3}{4} \Rightarrow -a^2(b + c) + abc = -\frac{3}{4} \]
Substituting \( b + c = \frac{1}{4} \):
\[ -a^2\left(\frac{1}{4}\right) + abc = -\frac{3}{4} \Rightarrow abc = -\frac{3}{4} + \frac{a^2}{4} \quad \text{(Equation 2)} \]

From \( S_4 \):
\[ -a^2bc = \frac{9}{8} \Rightarrow bc = -\frac{9}{8a^2} \quad \text{(Equation 3)} \]

### Step 4: Substitute and Solve for \( a^2 \)
Using Equation 1 in Equation 2:
\[ abc = -\frac{3}{4} + \frac{a^2}{4} \]
Substituting \( c = \frac{1}{4} - b \) into Equation 3:
\[ b\left(\frac{1}{4} - b\right) = -\frac{9}{8a^2} \]
This simplifies to:
\[ \frac{b}{4} - b^2 = -\frac{9}{8a^2} \]
Multiplying through by \( 8a^2 \):
\[ 2a^2b - 8a^2b^2 = 9 \]
Rearranging gives:
\[ 8a^2b^2 - 2a^2b + 9 = 0 \]

### Step 5: Solve the Quadratic Equation
Using the quadratic formula:
\[ b = \frac{-(-2a^2) \pm \sqrt{(-2a^2)^2 - 4 \cdot 8a^2 \cdot 9}}{2 \cdot 8a^2} \]
This leads to:
\[ b = \frac{2a^2 \pm \sqrt{4a^4 - 288a^2}}{16a^2} \]
Simplifying gives:
\[ b = \frac{2 \pm \sqrt{4(a^2 - 72)}}{16} \]
\[ b = \frac{1 \pm \sqrt{a^2 - 72}}{8} \]

### Step 6: Find Values of \( a \)
We need \( a^2 \geq 72 \) for real roots. Let \( a^2 = 3 \) (as derived earlier):
\[ b + c = \frac{1}{4}, \quad bc = -\frac{3}{8} \]
Solving gives:
\[ b = \frac{3}{4}, \quad c = -\frac{1}{2} \]

### Step 7: Final Roots
Thus, the roots of the equation are:
\[ \sqrt{3}, -\sqrt{3}, \frac{3}{4}, -\frac{1}{2} \]

### Final Answer
The roots of the equation \( 8x^4 - 2x^3 - 27x^2 + 6x + 9 = 0 \) are:
\[ \sqrt{3}, -\sqrt{3}, \frac{3}{4}, -\frac{1}{2} \]