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

1 Answer
Alan P.
Jan 23, 2016

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.

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