How do you find the discriminant and how many solutions does #x^2+4x+3=4# have?

2 Answers
Jun 8, 2018

Answer:

We get two real solutions

Explanation:

Writing your equation in the form
#x^2+4x-1=0#
the4n we get by the quadratic formula:

#x_{1,2}=2pm\sqrt(4+1)#
so we get
#x_1=2+sqrt(5)#

#x_2=2-sqrt(5)#

Jun 8, 2018

Answer:

#x = -2(1 +- sqrt5)#

Explanation:

#x^2 + 4x + 3 = 4#
#f(x) = x^2 + 4x - 1 = 0#
Discriminant D -->
#D = d^2 = b^2 - 4ac = 16 + 4 = 20# --> #d = +- 4sqrt5#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = -4/2 +- (4sqrt5)/2 = - 2 +- 2sqrt5#