How do you solve #x^2+2x-3>=0#?

1 Answer

Answer:

#x>=1, x<=-3#

Explanation:

The key to questions like this one is to solve it in the same way that you solve an equality (so one with an #=# sign), with two caveats:

  • that if you multiply or divide by a negative number, you flip the greater than sign, and
  • we'll be setting zones to see what is valid and what isn't.

So starting with the original:

#x^2+2x-3>=0#

we can factor this into:

#(x+3)(x-1)>=0#

and solve for x:

#x>=-3# and #x>=1#

Ok, so we know that we have 3 zones: #x>=1#, #x<=-3#, and the region between 1 and #-3#. Let's test #x=-2# to see if the middle region is valid:

#x^2+2x-3>=0#

#(-2)^2+2(-2)-3>=0#

#4-4-3>=0#

#-3>=0# - nope.

And let's test #x=-4# to test the left-most region:

#x^2+2x-3>=0#

#(-4)^2+2(-4)-3>=0#

#16-8-3>=0#

#5>=0# - yes.

And we can test the right-most region with #x=2#:

#x^2+2x-3>=0#

#2^2+2(2)-3>=0#

#4+4-3>=0#

#5>=0# - yes.

So the answer in notation is #x>=1, x<=-3#. The graph is below (on a number line, this will be solid lines on the x-axis in the shaded area starting from a solid point at x=1 and going right and another from x=-3 and going left.)

graph{x^2+2x-3>=0 [-10, 10, -1, 1]}