Question

using the quadratic formula to solve $11x^2 - 4x = 1$, what are the values of $x$?
$\frac{2}{11} \pm \frac{\sqrt{15}}{11}$
$\frac{2}{11} \pm \frac{2\sqrt{15}}{11}$
$\frac{2}{11} \pm \frac{\sqrt{5}}{11}$
$\frac{2}{11} \pm \frac{\sqrt{11}}{11}$

Explanation:

Step1: Rewrite the equation in standard form

The given equation is \(11x^{2}-4x = 1\). Subtract 1 from both sides to get it in the standard quadratic form \(ax^{2}+bx + c=0\). So, \(11x^{2}-4x - 1=0\), where \(a = 11\), \(b=-4\), and \(c=-1\).

Step2: Recall the quadratic formula

The quadratic formula is \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

Step3: Calculate the discriminant \(b^{2}-4ac\)

Substitute \(a = 11\), \(b=-4\), and \(c=-1\) into the discriminant formula. \(b^{2}-4ac=(-4)^{2}-4\times11\times(-1)=16 + 44=60\).

Step4: Substitute into the quadratic formula

Now, substitute \(a = 11\), \(b=-4\), and \(b^{2}-4ac = 60\) into the quadratic formula:
\(x=\frac{-(-4)\pm\sqrt{60}}{2\times11}=\frac{4\pm\sqrt{4\times15}}{22}=\frac{4\pm2\sqrt{15}}{22}\)
Simplify the fraction by dividing numerator and denominator by 2: \(x=\frac{2\pm\sqrt{15}}{11}=\frac{2}{11}\pm\frac{\sqrt{15}}{11}\)

Answer:

\(\frac{2}{11}\pm\frac{\sqrt{15}}{11}\) (corresponding to the first option \(\frac{2}{11}\pm\frac{\sqrt{15}}{11}\))