How do you solve the inequality #x^2+2x>=24#?

2 Answers
May 9, 2017

Answer:

By trial and error. Your answer is with the exception of -6<x<4, all x values provide solution.

Explanation:

x must be a real number and let me try +3 provide the solution. There are two numbers provide the solution (one is positive and the other one is negative):

#3^2+2*3=15# therefore it does not provide solution.

Now let me try +4

#4^2+2*4=24# Yes. This is one answer. Therefore x must be greater than or equal to +4.

On the negative side, let me try -3 first:

#(-3)^2-2*3=2# does not satisfy the requested condition.

Try -4 now:

#(-4)^2-2*4=8# does not satisfy the requested condition.

Try -5 now:

#(-5)^2-2*5=15# does not satisfy the requested condition.

Try -6 now:

#(-6)^2-2*6=24# Yes it satisfies.

Any number less than -6 (including -6) will satisfy the given.

For instance (-8):

#(-8)^2-2*8=48# which is greater than 24.

Now your solution is x must be greater than or equal to 4 or x must be less than or equal to -6.

May 9, 2017

Answer:

The solution is #x in (-oo,-6] uu[4,+oo)#

Explanation:

We solve this equation with a sign chart

Let's rearrange and factorise the inequality

#x^2+2x>=24#

#x^2+2x-24>=0#

#(x+6)(x-4)>=0#

Let #f(x)=(x+6)(x-4)#

Now, we construct the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-6##color(white)(aaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+6##color(white)(aaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaa)##+##color(white)(aaaa)##-##color(white)(aaaa)##+#

Therefore,

#f(x)>=0#, when #x in (-oo,-6] uu[4,+oo)#