Questions: Solve the equation: (16 r^2-8 r-11=0). Fully simplify your answer, including any non-real solutions. (r=)

Solution Steps

To solve the quadratic equation \(16r^2 - 8r - 11 = 0\), we can use the quadratic formula, which is given by \(r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 16\), \(b = -8\), and \(c = -11\). We will calculate the discriminant \(b^2 - 4ac\) to determine the nature of the roots and then apply the quadratic formula to find the solutions.

The given quadratic equation is \(16r^2 - 8r - 11 = 0\). Here, we identify the coefficients as follows:

• \(a = 16\)
• \(b = -8\)
• \(c = -11\)

We calculate the discriminant using the formula \(D = b^2 - 4ac\): \[ D = (-8)^2 - 4 \cdot 16 \cdot (-11) = 64 + 704 = 768 \]

Using the quadratic formula \(r = \frac{-b \pm \sqrt{D}}{2a}\), we find the roots: \[ r_1 = \frac{-(-8) + \sqrt{768}}{2 \cdot 16} = \frac{8 + \sqrt{768}}{32} \] \[ r_2 = \frac{-(-8) - \sqrt{768}}{2 \cdot 16} = \frac{8 - \sqrt{768}}{32} \]

Calculating the square root: \[ \sqrt{768} \approx 27.7128 \]

Now substituting back into the equations for \(r_1\) and \(r_2\): \[ r_1 = \frac{8 + 27.7128}{32} \approx \frac{35.7128}{32} \approx 1.1160 \] \[ r_2 = \frac{8 - 27.7128}{32} \approx \frac{-19.7128}{32} \approx -0.6160 \]

Final Answer