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