Solve the equation using the quadratic formula.
$2 x^2-13 x-1=0$
To solve the quadratic equation \(2x^2 - 13x - 1 = 0\), we first identify the coefficients \(a\), \(b\), and \(c\) from the standard quadratic form \(ax^2 + bx + c = 0\). Here, \(a = 2\), \(b = -13\), and \(c = -1\).
The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
First, we calculate the discriminant \(D = b^2 - 4ac\):
\[D = (-13)^2 - 4(2)(-1) = 169 + 8 = 177\]
The discriminant is 177, which is not a perfect square, so the roots will contain a radical.
Now, we substitute the values of \(a\), \(b\), and \(D\) into the quadratic formula:
\[x = \frac{-(-13) \pm \sqrt{177}}{2(2)}\]
Simplifying this expression, we get:
\[x = \frac{13 \pm \sqrt{177}}{4}\]
Thus, the two solutions to the equation are \(x = \frac{13 + \sqrt{177}}{4}\) and \(x = \frac{13 - \sqrt{177}}{4}\).
The solutions to the quadratic equation \(2x^2 - 13x - 1 = 0\) are:
\[x = \frac{13 + \sqrt{177}}{4}, \quad x = \frac{13 - \sqrt{177}}{4}\]
The solution set is \(\left\{ \frac{13 - \sqrt{177}}{4}, \frac{13 + \sqrt{177}}{4} \right\}\).
Quadratic equations often appear when calculating the trajectory of a launched projectile or determining the optimal dimensions of an area under specific constraints. For example, if you are designing a rectangular garden and know the relationship between its length and width such that the area equals a specific value, you might end up with a quadratic equation like this one. Solving for the dimensions ensures that the garden fits perfectly within the allocated space while meeting your design goals.