How do you find the roots, real and imaginary, of #y= x^2 - 3x - (2x+4)^2 # using the quadratic formula?

1 Answer
Mar 16, 2016

Answer:

real roots: #x=-16/3,-1#
imaginary roots: none

Explanation:

#1# Start by converting the equation into standard form.

#y=x^2-3x-(2x+4)^2#

#y=x^2-3x-(color(crimson)(2x)# #color(blue)(+4))(color(orange)(2x)# #color(teal)(+4))#

#y=x^2-3x-(color(crimson)(2x)(color(orange)(2x))# #color(crimson)(+2x)(color(teal)4)# #color(blue)(+4)(color(orange)(2x))# #color(blue)(+4)(color(teal)4))#

#y=x^2-3x-(4x^2+8x+8x+16)#

#y=x^2-3x-(4x^2+16x+16)#

#y=x^2-4x^2-3x-16x-16#

#y=color(brown)(-3)x^2# #color(turquoise)(-19)x# #color(violet)(-16)#

#2#. Identify the #color(brown)a,color(turquoise)b,# and #color(violet)c# values of the quadratic equation. Then plug the values into the quadratic formula to solve for the roots.

#color(brown)(a=-3)color(white)(XXXXX)color(turquoise)(b=-19)color(white)(XXXXX)color(violet)(c=-16)#

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

#x=(-(color(turquoise)(-19))+-sqrt((color(turquoise)(-19))^2-4(color(brown)(-3))(color(violet)(-16))))/(2(color(brown)(-3)))#

#x=(19+-sqrt(361-192))/-6#

#x=(19+-sqrt(169))/-6#

#x=(19+-13)/-6#

#x=(19+13)/-6color(white)(X),color(white)(X)(19-13)/-6#

#x=32/-6color(white)(X),color(white)(X)6/-6#

#color(green)(|bar(ul(color(white)(a/a)x=-16/3color(white)(i),-1color(white)(a/a)|)))#

#:.#, the real roots are #x=-16/3# or #x=-1#.