Questions: Find the roots of the equation (x^2+x+4)(x-3)^2=0 a. 3,3,-1/2+2 i,-1/2-2 i b. 3, 3, -1/2+sqrt(15)/2 i,-1/2-sqrt(15)/2 i C. 3,3, 1/2+sqrt(15)/2 i, 1/2-sqrt(15)/2 i d. 3,-1/2+sqrt(15)/2 i,-1/2-sqrt(15)/2 i

Solution Steps

To find the roots of the given equation \((x^{2}+x+4)(x-3)^{2}=0\), we need to solve each factor separately. The roots of the equation will be the values of \(x\) that make each factor equal to zero. For the factor \((x-3)^{2}=0\), the root is \(x=3\). For the quadratic factor \(x^{2}+x+4=0\), we can use the quadratic formula to find the complex roots.

The given equation is \((x^{2}+x+4)(x-3)^{2}=0\). To find the roots, we set each factor equal to zero.

For the factor \((x-3)^{2}=0\): \[ x - 3 = 0 \implies x = 3 \] This gives us a double root at \(x = 3\).

For the quadratic factor \(x^{2}+x+4=0\), we apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \(a = 1\), \(b = 1\), and \(c = 4\). The discriminant is calculated as: \[ b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 4 = 1 - 16 = -15 \] Since the discriminant is negative, the roots will be complex: \[ x = \frac{-1 \pm \sqrt{-15}}{2} = \frac{-1 \pm i\sqrt{15}}{2} \] This results in two complex roots: \[ x = -\frac{1}{2} + \frac{\sqrt{15}}{2}i \quad \text{and} \quad x = -\frac{1}{2} - \frac{\sqrt{15}}{2}i \]

Final Answer