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If a, b are the roots of x^(2)+x+1=0, th...

If a, b are the roots of `x^(2)+x+1=0`, then `a^(2)+b^(2)` is

A

ab

B

`a^(3)+b^(3)`

C

2ab

D

`a^(-2)+b^(-2)`

Text Solution

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The correct Answer is:
To solve the problem, we need to find the value of \( a^2 + b^2 \) given that \( a \) and \( b \) are the roots of the equation \( x^2 + x + 1 = 0 \). ### Step-by-step solution: 1. **Identify the roots of the equation**: The roots of the quadratic equation \( x^2 + x + 1 = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 1 \), and \( c = 1 \). \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \] This simplifies to: \[ x = \frac{-1 \pm i\sqrt{3}}{2} \] Thus, the roots are: \[ a = \frac{-1 + i\sqrt{3}}{2}, \quad b = \frac{-1 - i\sqrt{3}}{2} \] 2. **Calculate \( a^2 + b^2 \)**: We can use the identity: \[ a^2 + b^2 = (a + b)^2 - 2ab \] From Vieta's formulas, we know: - \( a + b = -1 \) (the coefficient of \( x \) with a negative sign) - \( ab = 1 \) (the constant term) Now, substituting these values: \[ a^2 + b^2 = (-1)^2 - 2 \cdot 1 = 1 - 2 = -1 \] 3. **Final result**: Therefore, the value of \( a^2 + b^2 \) is: \[ \boxed{-1} \]
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