## Algebra: A Combined Approach (4th Edition)

$x=8$
Given: $x^2-16x+64=0$ $a=1, \ b=-16, \ c=64$ Using the quadratic formula: $\dfrac{-b \ \pm \sqrt{b^2-4ac}}{2a}$ we have: $\dfrac{-(-16) \ \pm \sqrt{(-16)^2-4\times 1\times 64}}{2\times 1} = \dfrac{16 \ \pm \sqrt{256-256}}{2} = \dfrac{16 \ \pm \sqrt{0}}{2} = \dfrac{16}{2} = 8$ The only solution is $x=8$ because the discriminante is equal to $0$