To solve the problem, we need to find the roots of the polynomial equation:
\[ x^4 - 2x^3 - 3x^2 + 10x - 10 = 0 \]
given that the sum of two of the roots is zero. Let's denote the roots as \( a, -a, b, c \).
### Step 1: Define the roots
Let the roots be \( a, -a, b, c \). Since the sum of \( a \) and \( -a \) is zero, we can express the roots as:
\[ \text{Roots: } a, -a, b, c \]
### Step 2: Use Vieta's formulas
According to Vieta's formulas, for a polynomial of the form \( x^4 + px^3 + qx^2 + rx + s = 0 \), the relationships among the roots can be expressed as follows:
1. The sum of the roots \( (a + (-a) + b + c) = 0 \) (which is already satisfied).
2. The sum of the products of the roots taken two at a time \( (ab + ac + (-a)b + (-a)c + b(-a) + c(-a)) = -3 \).
3. The sum of the products of the roots taken three at a time \( (abc + (-a)ab + (-a)ac + b(-a)c) = 10 \).
4. The product of the roots \( (-a^2bc) = -10 \).
### Step 3: Set up equations
From the above relationships, we can derive the following equations:
1. \( b + c = 2 \) (since \( a + (-a) = 0 \))
2. \( bc - a^2 = -3 \) (from the second Vieta's relation)
3. \( -a^2bc = -10 \) (from the fourth Vieta's relation)
### Step 4: Solve for \( a^2 \) and \( bc \)
From equation 3, we can express \( bc \):
\[ bc = \frac{10}{a^2} \]
Substituting \( bc \) into equation 2 gives:
\[ \frac{10}{a^2} - a^2 = -3 \]
Multiplying through by \( a^2 \) to eliminate the fraction results in:
\[ 10 - a^4 = -3a^2 \]
Rearranging gives:
\[ a^4 - 3a^2 - 10 = 0 \]
### Step 5: Let \( y = a^2 \)
Let \( y = a^2 \). Then the equation becomes:
\[ y^2 - 3y - 10 = 0 \]
### Step 6: Solve the quadratic equation
Using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ y = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} \]
\[ y = \frac{3 \pm \sqrt{9 + 40}}{2} \]
\[ y = \frac{3 \pm \sqrt{49}}{2} \]
\[ y = \frac{3 \pm 7}{2} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{10}{2} = 5 \)
2. \( y = \frac{-4}{2} = -2 \) (not possible since \( y = a^2 \))
Thus, we have:
\[ a^2 = 5 \]
\[ a = \pm \sqrt{5} \]
### Step 7: Find \( b \) and \( c \)
Now substituting \( a^2 = 5 \) back into the equation for \( bc \):
\[ bc = \frac{10}{5} = 2 \]
And from \( b + c = 2 \), we can set up the equations:
1. \( b + c = 2 \)
2. \( bc = 2 \)
This leads to the quadratic equation:
\[ x^2 - 2x + 2 = 0 \]
### Step 8: Solve for \( b \) and \( c \)
Using the quadratic formula again:
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} \]
\[ x = \frac{2 \pm \sqrt{4 - 8}}{2} \]
\[ x = \frac{2 \pm \sqrt{-4}}{2} \]
\[ x = 1 \pm i \]
### Final Roots
Thus, the roots of the polynomial are:
\[ \sqrt{5}, -\sqrt{5}, 1 + i, 1 - i \]
### Summary of Roots
The roots are:
\[ \sqrt{5}, -\sqrt{5}, 1 + i, 1 - i \]
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