How do you find the roots, real and imaginary, of #y= 2x^2-15x-(4x+3)^2 # using the quadratic formula?

1 Answer
Jun 30, 2016

Answer:

#x~~-1.69" and "x~~-0.38" to 2 decimal places")#

Explanation:

You need to add like terms such that you and up with equation form:#" "y=ax^2+bx+c#

To do this expand the brackets and simplify.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Expanding the brackets")#

Note that #[-(4x+3)]^2" is not the same as "-[(4x+3)^2]#
The condition we have is#" "-[(4x+3)^2]#

#color(blue)((4x+3))color(brown)((4x+3))#

#color(brown)(color(blue)(4x)(4x+3)color(blue)(+3)(4x+3)) #

#16x^2+12x" "+12x+9" "=" "color(green)(16x^2+24x+9)#

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Substituting the expanded brackets and solving")#

#=>y=2x^2-15x-(color(green)(16x^2+24x+9))#

Multiply everything inside the brackets by -1 and group like terms

#=>y=2x^2-16x^2-15x-24x-9#

#=>y=-14x^2-29x-9#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From standard form we have:
#a=-14#
#b=-29#
#c=-9#

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

becomes#" "x=(+29+-sqrt((-29)^2-4(-14)(-9)))/(2(-14)#

#x=(29+-sqrt(841-504))/(-28)#

#x=(29+-sqrt(337))/(-28)" "larr 337" is a prime number"#

#color(blue)(x~~-1.69" and "x~~-0.38" to 2 decimal places")#

Tony B