How do you find the zeros, real and imaginary, of #y=-4x^2-4x-15# using the quadratic formula?

1 Answer
Nov 17, 2015

Answer:

Finding the values of #x# for #color(white)(xxx)y=-4x^2-4x-15 = 0#

#x!in RRcolor(white)(xx)# but #color(white)(xx) x inCC -> x=-1/2 +-isqrt(7/2)#

Explanation:

Tont B

Using standard for of #y=ax^2+bx+c#

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

Let:
#a=-4#
#b=-4#
#c=-15#

So by substitution we have:

#x= (-(-4) +- sqrt((-4)^2 -(4)(-4)(-15)))/(2(-4)#

#x=(+4 +- sqrt(16-240))/(2(-4))#

#x=(4+-sqrt(-224))/(-8)#

partitioning 224 into prime numbers and squaring where able

#x= (4+- sqrt((-1) times 2^2 times 2^2 times 2 times 7))/(-8)#

#x= (4+-4sqrt(-14))/(-8)#

#x=(1+- i sqrt(14))/(-2)#
As we have #+-# in the numerator before the root having a negative denominator makes no tangible difference to that root. However, it does have an effect on the 1 preceding it. Consequently we have #-1/2#.

#x=-1/2 +- isqrt(14/4)#

#x=-1/2 +-isqrt(7/2)#