How do you find the roots, real and imaginary, of #y=2x^2 + 13x +6 # using the quadratic formula?

1 Answer
Mar 16, 2016

Answer:

real roots: #x=-1/2,-6#
imaginary roots: none

Explanation:

#1#. Since the given equation is already in standard form, identify the #color(blue)a,color(darkorange)b,# and #color(violet)c# values. Then plug the values into the quadratic formula to solve for the roots.

#y=color(blue)2x^2# #color(darkorange)(+13)x# #color(violet)(+6)#

#color(blue)(a=2)color(white)(XXXXX)color(darkorange)(b=13)color(white)(XXXXX)color(violet)(c=6)#

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

#x=(-(color(darkorange)(13))+-sqrt((color(darkorange)(13))^2-4(color(blue)2)(color(violet)(6))))/(2(color(blue)2))#

#x=(-13+-sqrt(169-48))/4#

#x=(-13+-sqrt(121))/4#

#x=(-13+-11)/4#

#x=(-13+11)/4color(white)(X),color(white)(X)(-13-11)/4#

#x=-2/4color(white)(X),color(white)(X)-24/4#

#color(green)(|bar(ul(color(white)(a/a)x=-1/2,-6color(white)(a/a)|)))#

#:.#, the real roots are #x=-1/2# or #x=-6#.