#-16x^2+8x-1=0#?

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Answer 1 · 2018-04-18T10:53:16

#x=1/4 or 0.25#

So this is a quadratic with only 1 solution. I personally like to use the quadratic formula. Apply the following:

We dont have to re-write our numbers as they already follow the order we need them to be in.

So:

#a=-16#
#b=8#
#c=-1#

Now put these into this formula which is:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

So we get: #x=(-8+-sqrt(8^2(-4*-16*-1)))/(2*-16)#

Now we simplify to get: #x=(-8+-sqrt(64-64))/(-32)#

Simplify further to get: #x=(-8)/(-32)#

Now divide by 8 on the denominator and the numerator to get:

#(-1)/(-4)#

Which is the same thing as #1/4# which is 0.25 but I personally like to keep them as fractions.

Hope this helps =)

Answer 2 · 2018-04-19T07:24:47

#x=1/4#

In this quadratic, the negative at the beginning is not comfortable.

Divide the whole equation by #-1#

#-16x^2 +8x-1=0 " "rarr" "16x^2-8x+1=0#

This can factorise:
Find factors of #16# which add to #8#.
Both brackets will have a negative sign.

#(4x-1)(4x-1)=0#

#4x-1=0#

#4x=1#

#x=1/4#