Answer:
2b) x < - 3
2e) x < 3 or x < -3
Explanation:
Format for quadratic equations (the only difference in quadratic equations is that the equal sign is replaced with any of the inequality symbols (< , > , ≥ , and ≤)
ax² + bx + c = 0
- To factor quadratic inequalities (we're using 2b) as a demonstration)
1) Find two numbers that make the product equal to a×c (1×9) which is 9 and the sum that equals to b which is 6. Those two numbers in this case is positive 3 and positive 3
2) Spilt the middle term with those two numbers which are positive 3 and positive 3 so the inequality x² + 6x + 9 is going to become x² + 3x + 3x + 9
3) Take the common terms and simplify the common terms in this case the common terms are x and + 3 so it can be factored as (x + 3) and (x + 3)
To find each x make each term greater than 0 so it would be x + 3 < 0 and x + 3 < 0
4) Solve for x (it is the same way to solve x in inequalities as it were for equations which is doing inverse operations to order to islolate x) It would be x + 3 - 3 < 0 - 3 which is x < - 3
2e)
1) You essentially do the same steps for solving this question too but a shortcut way of doing factoring is to do the perfect square method which is to look a c which is 9 and if its a perfect squares you can factor it as (x+3) and (x-3) since not only is the 9 negative both of these terms when added cancel out each other.
2)To find x make everything < 0 so it would be x + 3 < 0 and x - 3 < 0
3) Solve for x it would be x + 3 - 3 < 0 - 3 and x - 3 + 3 < 0 + 3 . The roots are x < - 3 or x < 3
I hope this helps :)