Find the x- intercepts(see picture). I used the quadratic formula and I substitute. I found 2 intercepts but I am not sure. Can you make it more clear for me? Thanks!

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Answer 1 · 2018-01-22T22:31:56

#-1+sqrt6/6# and #-1-sqrt6/6#
(note that the answer asks for exact form)

the quadratic formula should work, since it has given you two distinct intercepts.

#6x^2+12x+5 = 0#
#ax^2 + bx+c = 0#

#a =6, b = 12, c= 5#

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

#x = (-12 +- sqrt(12^2-(4*6*5)))/(12)#

#= (-12 +- sqrt(144-120))/(12)#

#= (-12 +- sqrt(24))/(12)#

[note: the fact that #b^2-4ac# is above #0# means that there are two distinct real roots.]

#sqrt(24) = (sqrt4) * (sqrt6) = 2sqrt6#

#(-12 +- sqrt(24))/(12) = (-6+-sqrt6)/(6)#

#= -1 +- sqrt6/6#

this gives us the solutions #-1+sqrt6/6# and #-1-sqrt6/6#.

desmos,com/calculator

this graph shows that the parabola does meet the #x-#axis twice.
the graph has #2# #x-#intercepts.

the graph shows #x-#values of #-1.4082# and #-0.5918#,

which are decimal approximations of #-1-sqrt6/6# and #-1+sqrt6/6# respectively.