How do you find the roots, real and imaginary, of #y= 2(x+1)^2-(x-4)^2 # using the quadratic formula?

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Answer 1 · 2017-09-04T10:49:07

#x = -6 + 5sqrt(2) " or " x = -6 - 5sqrt(2)#

Need to expand out the brackets:

#y = 2(x^2 + 2x + 1) - (x^2 - 8x + 16)#

#y = x^2+12x - 14#

The roots of a quadratic of the form

#ax^2 + bx + c #

given by

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

In this case #a = 1, b = 12, c = -14#

#x = (-12 +- sqrt(144 -4(1)(-14)))/2 = (-12+-sqrt(144+56))/2 #

#x = (-12+-sqrt(200))/2#

#x = (-12 +- sqrt(100)sqrt(2))/2 = (-12+-10sqrt(2))/2#

#therefore x = -6 + 5sqrt(2) " or " x = -6 - 5sqrt(2)#