Is it possible to factor #y=4x^2 - 8x + 3 #? If so, what are the factors?

1 Answer
Apr 6, 2016

Answer:

#x= 3/2 or 1/2#

Explanation:

The coefficient of #x^2# is positive so the graph is of format type #uu# so the vertex is a minimum.

Standard equation for:# y=ax^2+bx+c#

Where #a=4"; "b=-8"; "c=+3#

and #" " x=(-b+-sqrt(b^2-4ac))/(2a)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#x= (+8+-sqrt((-8)^2-4(4)(3)))/( 2(4))#

#x= 1+-sqrt(16 )/8#

#x= 3/2 or 1/2#