How do you solve #3z ^ { 2} + 4z - 7= 0#?

2 Answers

Answer:

#z=-7/3 or z = 1#

Explanation:

Using the quadratic equation

#z_(1,2) = (-b +- sqrt(b^2 –4ac))/(2a)#

you get

#z= (-4 +- sqrt( 4^2 -4xx3xx (-7)))/(2xx3)#

#z= (-4 +- sqrt(16+84))/6#

#z= (-4 +- sqrt(100))/6#

#z= (-4 +- 10)/6#

#z= -14/6 or z = 6/6#

#z=-14/6 or z = 1#

#z= -7/3 or z = 1#

Apr 4, 2018

Answer:

1, and #-7/3#

Explanation:

#3z^2 + 4z - 7 = 0#
Use shortcut when a + b + c = 0
#z = 1#, and #z = c/a = - 7/3#

Reminder of shortcut --> #y = ax^2 + bx + c = 0#

  1. When a + b + c = 0
    #x1 = 1#, and #x2 = c/a#
    Example: #y = 3x^2 + 4x - 7 = 0#
    #x1 = 1#, and #x2 = - 7/3#

  2. When a - b + c = 0.
    x1 = - 1, and #x2 = - c/a#
    Example. #y = 5x^2 + 3x - 2x = 0#
    #x1 = - 1#, and #x2 = -c/a = 2/5#