2x^3-x^2-1=0
To solve the equation \(2x^3 - x^2 - 1 = 0\), we can use numerical methods, graphing, or algebraic methods to find its roots. Here are steps on how to approach it:
Finding Rational Roots (Rational Root Theorem): We can check for possible rational roots using the Rational Root Theorem. The possible rational roots are factors of the constant term (-1) divided by the leading coefficient (2):
Possible rational roots: \( \pm 1, \pm \frac{1}{2} \)
Testing Possible Roots: We will test these possible roots in the polynomial.
- Testing \(x = 1\): \[ 2(1)^3 - (1)^2 - 1 = 2 - 1 - 1 = 0 \] So, \(x = 1\) is a root.
Factoring the Polynomial: Given that \(x = 1\) is a root, we can use synthetic division or polynomial long division to factor the polynomial:
Dividing \(2x^3 - x^2 - 1\) by \(x - 1\):
Using synthetic division: \[ \begin{array}{r|rrr} 1 & 2 & -1 & 0 & -1 \\ & & 2 & 1 & 1 \\ \hline & 2 & 1 & 1 & 0 \\ \end{array} \]
The result is \(2x^2 + x + 1\).
Therefore, we can factor the original equation as: \[ 2x^3 - x^2 - 1 = (x - 1)(2x^2 + x + 1) \]
Finding Other Roots: Now, we need to solve \(2x^2 + x + 1 = 0\). We can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 2\), \(b = 1\), and \(c = 1\).
\[ x = \frac{-1 \pm \sqrt{(1)^2 - 4(2)(1)}}{2(2)} = \frac{-1 \pm \sqrt{1 - 8}}{4} = \frac{-1 \pm \sqrt{-7}}{4} \]
This gives us complex roots: \[ x = \frac{-1 \pm i\sqrt{7}}{4} \]
Final Roots: Thus, the complete set of roots for the equation \(2x^3 - x^2 - 1 = 0\) is: \[ x = 1, \quad x = \frac{-1 + i\sqrt{7}}{4}, \quad x = \frac{-1 - i\sqrt{7}}{4} \]
In summary, the roots of the equation \(2x^3 - x^2 - 1 = 0\) are:
- \(x = 1\)
- \(x = \frac{-1 + i\sqrt{7}}{4}\)
- \(x = \frac{-1 - i\sqrt{7}}{4}\)
