Question

find the solution set of this inequality. x(x+1) ≤ 20

so i have to take out the 20 first ?

Answer 1

Try turning it into an inequality with 0 on one side.

Answer 2

solution please! :D

Answer 3

ok, add (-20) to both sides of the inequality, that yields
x(x + 1) - 20 ≤ 0
x^2 + x - 20 ≤ 0
(x + 5)(x - 4) ≤ 0

Could you take it from there? :)

Answer 4

what do you mean by "take out the 20 first"?

this is the way I would proceed

\[x(x+1)\leq20\]
first determine where the expression is equal to zero, these are the"x" values where the statement will change from true to false or visa versa
so I want to solve for x in
\[x(x+1)=20\]
\[x^2+x-20=0\]
factoring
\[(x+5)(x-4)=0\]

so \[x=-5\] or \[x=4\]

now test the original statement for intervals of truth\[(x+5)(x-4)\leq 0\]

as test numbers I choose \[-6, 0, and, 5\]
-6 results in a false statement
0 results in a true statement
and 5 results in a false statement

so the solution set is the closed interval\[[-5,4]\]

Answer 5

thank you!

Answer 6

oh yeah i forgot 5&4. geez