How do you solve #x^2+4x= -1#?

1 Answer
May 20, 2016

Answer:

#x = -2 \pm \sqrt{3}#

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 #x^2 + 4x = -1#.

  • 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: #x = -2 \pm \sqrt{3}#