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)
