How do you solve the quadratic using the quadratic formula given #2a^2-6a-3=0# over the set of complex numbers?

1 Answer
SCooke
May 13, 2017

The solution does not require complex numbers.

Explanation:

You simply put the coefficients of the equation into the formula and solve it algebraically.
For #a^x2 + b*x + c = 0#, the values of x which are the solutions of the equation are given by:

#x = ​(−b±√[​b​^2​​−4ac])/(2a)#

In this case, a = 2, b = -6 and c = -3

#x = ​−(-6) ± sqrt[((-6)^​2)​ ​− (4*2*-3))/(2*2)#

#x = ​6 ± sqrt[(36 + 24])/4# ; #x = ​(6 ± sqrt[60])/4#
#x = ​(6 ± 7.75)/4#

#x = -0.4375, and x = 3.4375#

CHECK : 2(-0.4375^2) – 6(-0.4375) -3 = 0 ; 2*(0.19) + 2.62 – 3 = 0 ;
0.38 + 2.62 – 3 = 0 ; 0 = 0 correct.
See http://www.purplemath.com/modules/quadform.htm for instructions.

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