Question

How can you describe the solution set of the equation 16x^2 - 8x + 1 = 0?

Answer 1

16x^2 - 8x + 1 = 0
16x^2 - 4x - 4x + 1 = 0
4x (4x - 1) - 1 (4x - 1) = 0
(4x - 1) (4x - 1) = 0
(4x - 1)^2 = 0

4x = 1 
x = 1/4

Answer 2

$16x^2 - 8x + 1 = 0 \\ \\ \boxed{\Delta=b^2-4ac} \\ \\ \Delta=(-8)^2-4\cdot16\cdot1=0 \\ ==============\\ \displaystyle \\ \Delta\ \textgreater \ 0 \\ \\ X_{1,2}= \frac{-b\pm \sqrt{\Delta} }{2a} \\ \\ --------- \\ \Delta=0 \\ \\ X1=X2= \frac{-b}{2a} \\ \\ ---------- \\ \\ \Delta\ \textless \ 0 \\ \\ X_{1,2}= \frac{-b\pm i \sqrt{\Delta} }{2a} \\ \\ ----------- \\ \\ \text{We have delta=0} \\ \\ X1=X2= \frac{\not8}{2\cdot\not16} = \frac{1}{2\cdot2} = \bold{\bold{\frac{1}{4} }}$