How do you solve #x^2+4x= 1#?
1 Answer
Answer:
Explanation:
There are several ways to do it, by I would prefer in this case to use the completing the square method to find the root of the quadratic

In completing the square, we need to first make sure whether the a value of the quadratic equation equals to 1. If not, then you need to divide every coefficient and constant by the
#a# value to make sure that the#a# value is 1. In this case, this is not required. 
Now we need to divide the
#b# term by 2 and then square it. Do this to both sides.
#x^2 + 4x + (\frac{4}{2})^2 = 1 + (\frac{4}{2})^2#
#x^2 + 4x + 4 = 1 + 4#
#x^2 + 4x + 4 = 3# 
Put
#x^2 + 4x + 4# into a binomial.
#(x + 2)^2 = 3# 
To solve for
#x# , find the square root of both sides.
#\sqrt{(x+2)^2} = \sqrt{3}#
# x + 2 = \pm \sqrt{3}#
Notice the#\pm# in front of the square root of 3. This is because the square root of 3 can be both a positive and a negative (a negative times a negative is a positive). 
Isolate
#x# by subtracting 2 from both sides.
# x = 2 \pm \sqrt{3}#
Answer: