How do you find the zeros, real and imaginary, of #y= -7x^2+4x+2 # using the quadratic formula?
1 Answer
May 24, 2017
Explanation:
#y = -7x^2+4x+2#
is in the form:
#y = ax^2+bx+c#
with
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)#
