Question

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

There are two different real solutions because the expression has two different factors.

There are two different real solutions because the expression has three terms.

There is one real solution because the expression’s factors are the same.

There are no real solutions because the expression cannot be factored.

Answer 1

`16x^2 - 8x + 1 = 0 \\ \\ \boxed{\Delta=b^2-4ac} \\ \\ \Delta=(-8)^2-4\cdot16\cdot1=0 \\ ==============\\ \displaystyle \\ \Delta\ \textgreater \ 0 \\ \\ X_(1,2)= (-b\pm √(\Delta) )/(2a) \\ \\ --------- \\ \Delta=0 \\ \\ X1=X2= (-b)/(2a) \\ \\ ---------- \\ \\ \Delta\ \textless \ 0 \\ \\ X_(1,2)= (-b\pm i √(\Delta) )/(2a) \\ \\ ----------- \\ \\ \text{We have delta=0} \\ \\ X1=X2= (\\ot8)/(2\cdot\\ot16) = (1)/(2\cdot2) = \bold{\bold{(1)/(4) }}`

Answer 2

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