How do you solve #4x^2+40x+280=0# by completing the square?

1 Answer
Jul 2, 2017

Answer:

You can't, unless you want to have complex numbers. If you want complex numbers, then the answer would be #x=-5\pm3i\sqrt{5}#

Explanation:

You can simplify the quadratic to make completing the square easier:
#4x^2+40x+280=0#
#x^2+10x+70=0#

Completing the square:
#x^2+10x+70=0#
#x^2+10x=-70#
#x^2+10x+(10/2)^2=-70+(10/2)^2#
#x^2+10x+25=-70+25#
#(x+5)^2=-45#

We cannot continue here, since there is no way for the above equation to be true.

However, if you want to have complex numbers, we can continue:
#x+5=\pm\sqrt{-45}#
#x+5=\pmi\sqrt{45}#
#x+5=\pm3i\sqrt{5}#
#x=-5\pm3i\sqrt{5}#