Home > Algebra > Quadratic Formula

How do you find the zeros, real and imaginary, of #y=--x^2 +6x +8# using the quadratic formula?

Answer 1

Eli Assouline

#color(purple)(x = (3 - sqrt78/2), 3 + sqrt78/2#

#y = - x^2 + 6x = 8#

#a = -1, b = 6, c = 8#

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

#x =( -6 +- sqrt(6^2 + 32)) / -2#

#color(purple)(x = (3 - sqrt78/2), 3 + sqrt78/2#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Jace Anderson

To find the zeros of the quadratic function y = -x^2 + 6x + 8 using the quadratic formula, first identify the coefficients: a = -1, b = 6, and c = 8. Then apply the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a). Substituting the values, you get: x = (-(6) ± √((6)^2 - 4(-1)(8))) / (2(-1)). Simplify inside the square root, then solve for the real and imaginary zeros accordingly.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.