1x^2 - 8x + 16 = 0

a = 1 b = - 8 C = 16

1 answer

To solve the quadratic equation \(1x^2 - 8x + 16 = 0\) using the quadratic formula, we can identify the coefficients:

  • \(a = 1\)
  • \(b = -8\)
  • \(c = 16\)

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

First, calculate the discriminant (\(b^2 - 4ac\)):

\[ b^2 = (-8)^2 = 64 \] \[ 4ac = 4 \cdot 1 \cdot 16 = 64 \] \[ b^2 - 4ac = 64 - 64 = 0 \]

Since the discriminant is 0, there will be one real solution (a repeated root):

Now substitute the values into the quadratic formula:

\[ x = \frac{-(-8) \pm \sqrt{0}}{2 \cdot 1} \] \[ x = \frac{8 \pm 0}{2} \] \[ x = \frac{8}{2} = 4 \]

Thus, the equation \(1x^2 - 8x + 16 = 0\) has one repeated solution:

\[ \boxed{4} \]