How do you find the roots, real and imaginary, of #y=4x^2+12x-25# using the quadratic formula?

1 Answer
Mar 7, 2018

Answer:

#x=(-3+-sqrt34)/2#

Explanation:

#4x^2+12x-25=ax^2+bx+c#

set #4x^2+12x-25# to 0 to find the roots real and imaginary

Using the discriminant #b^2-4ac# to find out if there are imaginary roots

#(12)^2-4(4)(-25)=544#

#Delta("the discriminant")>0# So that means it have 2 unique real roots

Using the quadratic formula

#x=(-(12)+-sqrt((12)^2-4(4)(-25)))/(2(4))#

#x=(-12+-sqrt544)/8#

simplify the radical using #root(n)xy=root(n)x*root(n)y#

#x=(-12+-sqrt(16*34))/8#

#x=(-12+-sqrt(4)sqrt(34))/8#

#x=(-12+-2sqrt(34))/8#

Find the greatest common factor

#x=(-3+-sqrt(34))/2#

#x=(-3+sqrt(34))/2# and #-(3+sqrt(34))/2#

Using #C.T.S.# (complete the square)
To get #x^2# by itself divide the whole equation by #4#

#(4x^2+12x-25)/4=0/4#

#(cancel4x^2)/cancel4+(3cancel12x)/(1cancel4)-25/4=0#

#x^2+3xcancel(-25/4+25/4)=0+25/4#

#x^2+3x=25/4#

complete the square by dividing the coefficient of x by 2 and then squaring it

#(3/2)^2=9/4#

add this term to both sides

#x^2+3x+9/4=25/4+9/4#

#sqrt((x+3/2)^2)=+-sqrt(34/4#

#xcancel(+3/2-3/2)=(+-sqrt(34))/2+(-3)/2#

#x=(-3+-sqrt34)/2#

#:.# #x=(-3+sqrt34)/2# and #(-3-sqrt34)/2#