How do you find the zeros, real and imaginary, of #y=-7x^2-2x+3# using the quadratic formula?

1 Answer
Jan 23, 2016

Answer:

There are two real zeroes:
#color(white)("XXX")x=-4/7+sqrt(22)/7color(white)("XX")# and #color(white)("XX")x=-4/7-sqrt(22)/7#

Explanation:

The quadratic formula for an expression of the form
#color(white)("XXX")ax^2+bx+c#
tells us that the zeroes occur at
#color(white)("XXX")x=(-b+-sqrt(b^2-4ac))/(2a)#

For the given expression: #-7x^2-2x+3#
#color(white)("XXX")a=-7#
#color(white)("XXX")b=-2#
#color(white)("XXX")c=+3#

So the zeroes are at
#color(white)("XXX")x=(-(-2)+-sqrt((-2)^2+4(-7)(3)))/(2(-7))#
#color(white)("XXX")=(4+-sqrt(4+84))/(-14)#
#color(white)("XXX")=-(4+-2sqrt(22))/14#
#color(white)("XXX")=-2/7+-sqrt(22)/7#

Note that since the argument of the square root is positive,
the zeroes are Real.