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

2 Answers
Meave60
Jul 8, 2017

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)

Nghi N
Jul 8, 2017

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.

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