How do you find the zeros, real and imaginary, of #y=- 7x^2-15x -35# using the quadratic formula?

2 Answers
Jul 8, 2017

Answer:

#x=(15+isqrt(755))/(-14),##(15-isqrt(755))/(-14)#

Explanation:

The zeros of a quadratic equation are the values for #x#, and are where the parabola crosses the x-axis.

#y=-7x^2-15x-35#

Substitute a zero for the #y#.

#0=-7x^2-15x-35# is a quadratic equation in the form: #ax^2+bx+c=0#, where #a=-7#, #b=-15#, #c=-35#.

Use the quadratic formula to solve for #x# (the zeros).

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

Substitute the given values for #a, b, and c# into the formula.

#x=(-(-15)+-sqrt((-15)^2-4*-7*-35))/(2*-7)#

Simplify.

#x=(15+-sqrt(-755))/(-14)#

#x=(15+isqrt(755))/(-14),##(15-isqrt(755))/(-14)#

Jul 8, 2017

Answer:

2 imaginary roots:
#x = (-15 +- sqrt755)/14#

Explanation:

#y = - (7x^2 + 15x + 35) = 0#
Use new improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 225 - 980 = - 755 = 755i^2# -->
#d = +- isqrt755#
Because D < 0, there are 2 imaginary roots.
#x = - b/(2a) +- d/(2a) = - 15/14 +- isqrt755/14 = (-15 +- isqrt755)/14#

Note. Using the improved quadratic formula, the proceeding is much simpler, and the numeric computation is much easier.