How do you find the roots, real and imaginary, of #y=7x^2-2x + 6# using the quadratic formula?

Question

Abigail Allen · Accepted Answer

# (1 +- 6.4i)/7# #y = 7x^2 - 2x + 6 = 0# Use the improved quadratic formula, also called the quadratic formula in intercept form (Google, Yahoo, Bing Search). #D = d^2 = b^2 - 4ac = 4 - 168 = -164# --> #d = +- 12.80( i )# Since D < 0, there are no real roots. There are 2 complex roots --> #x = -b/(2a) +- d/(2a) = 2/14 +- (12.80i)/14 = (1 +- 6.4i)/7#

Ethan Adkins · Answer

undefined To find the roots, both real and imaginary, of the quadratic equation $ y = 7x^2 - 2x + 6 $ using the quadratic formula, follow these steps:

1. Identify the coefficients of the quadratic equation: $ a = 7 $, $ b = -2 $, and $ c = 6 $.

2. Substitute these values into the quadratic formula: $ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $.

3. Calculate the discriminant ($ b^2 - 4ac $): 
   $ b^2 - 4ac = (-2)^2 - 4 * 7 * 6 = 4 - 168 = -164 $.

4. Since the discriminant is negative ($ -164 $), the roots are imaginary.

5. Substitute the values of $ a $, $ b $, and the discriminant into the quadratic formula:
   $ x = \frac{-(-2) \pm \sqrt{-164}}{2 * 7} $.

6. Simplify the expression:
   $ x = \frac{2 \pm \sqrt{164}i}{14} $.

7. Further simplify by factoring out $ \sqrt{164} $:
   $ x = \frac{2 \pm 2\sqrt{41}i}{14} $.

8. Finally, divide both the real and imaginary parts by 2 to simplify:
   $ x = \frac{1 \pm \sqrt{41}i}{7} $.

Therefore, the roots of the quadratic equation $ y = 7x^2 - 2x + 6 $ are $ x = \frac{1 + \sqrt{41}i}{7} $ and $ x = \frac{1 - \sqrt{41}i}{7} $, where $ i $ represents the imaginary unit.