How do you find the zeros, real and imaginary, of #y= -7x^2+4x+2 # using the quadratic formula?

1 Answer
May 24, 2017

Answer:

#x = 2/7+3/7sqrt(2)" "# or #" "x = 2/7-3/7sqrt(2)#

Explanation:

#y = -7x^2+4x+2#

is in the form:

#y = ax^2+bx+c#

with #a=-7#, #b=4# and #c=2#

This has zeros given by the quadratic formula:

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

#color(white)(x) = (-color(blue)(4)+-sqrt(color(blue)(4)^2-4(color(blue)(-7))(color(blue)(2))))/(2(color(blue)(-7)))#

#color(white)(x) = (-4+-sqrt(16+56))/(-14)#

#color(white)(x) = (-4+-sqrt(72))/(-14)#

#color(white)(x) = (-4+-sqrt(6^2*2))/(-14)#

#color(white)(x) = (-4+-6sqrt(2))/(-14)#

#color(white)(x) = (-2+-3sqrt(2))/(-7)#

#color(white)(x) = 2/7+-3/7sqrt(2)#

That is:

#x = 2/7+3/7sqrt(2)" "# or #" "x = 2/7-3/7sqrt(2)#