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

1 Answer
Nov 17, 2015

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)