How do you solve #7x^2 - 4x - 1 = 0# using the quadratic formula?

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Answer 1 · 2016-04-06T07:42:31

#y=(2+sqrt11)/7# or #y=(2-sqrt11)/7#

Using quadratic formula the solution of #ax^2+bx+c=0# is given by

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

As in #7x^2-4x-1=0#, #a=7#, #b=-4# and #c=-1#,

solution of #7x^2-4x-1=0# is given by

#y=(-(-4)+-sqrt((-4)^2-4*7*(-1)))/(2*7)#

or #y=(4+-sqrt(16+28))/14=(4+-sqrt44)/14#

Note that #sqrt44=2sqrt11#, hence

#y=(4+-2sqrt11)/14=(2+-sqrt11)/7# i.e.

#y=(2+sqrt11)/7# or #y=(2-sqrt11)/7#

Answer 2 · 2016-04-06T09:14:32

#x=(4+sqrt11)/7,(4-sqrt11)/7#

#color(blue)(7x^2-4x-1=0#

This is a Quadratic equation (in form #ax^2+bx+c=0#)

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

Remember that #a,bandc # are the coefficients

So,

#color(purple)(a=7,b=-4,c=-1#

#rarrx=(-(-4)+-sqrt(-4^2-4(7)(-1)))/(2(7))#

#rarrx=(4+-sqrt(16-(-28)))/(14)#

#rarrx=(4+-sqrt(16+28))/(14)#

#rarrx=(4+-sqrt(44))/(14)#

#rarrx=(4+-sqrt(4*11))/(14)#

#rarrx=(4+-2sqrt(11))/(14)#

#rarrx=(cancel4^2+-cancel2^1sqrt(11))/(cancel14^7)#

#rarrx=(4+-sqrt11)/7#

Now we have two solutions

#color(indigo)(x=(4+sqrt11)/7#

#color(violet)(x=(4-sqrt11)/(7)#

#color(blue)(ul bar |x=(4+sqrt11)/7,(4-sqrt11)/7|#