Question

find the solutions of the quadratic equation $11x^2 - 6x - 2 = 0$. choose 1 answer: a $\frac{3}{11} pm \frac{sqrt{31}}{11}$ b $-\frac{3}{11} pm \frac{sqrt{31}}{11}$ c $\frac{3}{11} pm \frac{sqrt{31}}{11}i$ d $\frac{3}{11} pm \frac{sqrt{52}}{11}i$

Explanation:

Step1: Recall Quadratic Formula

For a quadratic equation \(ax^{2}+bx + c = 0\), the solutions are given by \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). Here, \(a = 11\), \(b=-6\), \(c = - 2\).

Step2: Calculate Discriminant (\(D\))

Discriminant \(D=b^{2}-4ac\). Substitute \(a = 11\), \(b=-6\), \(c=-2\):
\(D=(-6)^{2}-4\times11\times(-2)=36 + 88=124\)? Wait, no, wait: Wait, \(b=-6\), so \(b^{2}=(-6)^{2}=36\), \(4ac = 4\times11\times(-2)=-88\), so \(D=36-4\times11\times(-2)=36 + 88 = 124\)? Wait, no, wait the equation is \(11x^{2}-6x - 2=0\), so \(a = 11\), \(b=-6\), \(c=-2\). Then \(D=b^{2}-4ac=(-6)^{2}-4\times11\times(-2)=36 + 88 = 124\)? Wait, but \(124 = 4\times31\), so \(\sqrt{D}=\sqrt{124}=\sqrt{4\times31}=2\sqrt{31}\).

Step3: Apply Quadratic Formula

\(x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-(-6)\pm2\sqrt{31}}{2\times11}=\frac{6\pm2\sqrt{31}}{22}\). Simplify numerator and denominator by dividing by 2: \(\frac{3\pm\sqrt{31}}{11}=\frac{3}{11}\pm\frac{\sqrt{31}}{11}\). Also, since \(D = 124>0\), the solutions are real, so no imaginary unit \(i\).

Answer:

A. \(\frac{3}{11} \pm \frac{\sqrt{31}}{11}\)