How do you find the roots, real and imaginary, of #y= x^2 - 3x + 44-(3x-1)^2 # using the quadratic formula?

1 Answer
Aug 9, 2017

Answer:

You should know the 2 formulas:
1.) #(a - b)^2 = a^2+b^2-2ab#
2.) Quadratic Formula: #(-b+-sqrt(b^2-4ac))/(2a)#

Explanation:

Step 1.) Open the bracket term in question by formula no.1 and solve it further:

#y=x^2-3x+44-(3x-1)^2#
#y=x^2-3x+44-(9x^2+1-6x)#

Now solvng it further:
#y=x^2-3x+44-9x^2-1+6x#
#y=-8x^2+3x+43#

Step 2.) Now applying Quadratic formula:

#(-b+-sqrt(b^2-4ac))/(2a)#
#x= (-3+-sqrt(3^2-4(43)(-8)))/(2(-8)#
#x= (-3+-sqrt(9+1376))/(-16)#
#x= (-3+-sqrt(1385))/(-16)#
#x= (-3+-37.21)/(-16)#

Case 1:
when the sign is positive in between:
#x= (-3+37.21)/(-16)#
#x= (34.21)/(-16)#
#x=- 2.13#

Case 2:
when the negative is positive in between:
#x= (-3-37.21)/(-16)#
#x= (-40.21)/(-16)#
#x= 2.51#

That's all, Thank You.