Question 629494
Actually, your "one piece" is not right. I suspect you multiplied or divided both sides by a negative at some point and forgot the special rule that tells you to reverse the inequality.<br>
There are several ways to figure this out. One of them would be...
Make one side of the inequality zero. The fast way to do this would be to subtract 5 from each side. But I prefer to avoid the special rule. So I like to make sure the variable term has a positive coefficient. So I am going to subtract 8 and add {{{x^2}}} on both sides:
{{{0 < x^2-3}}}
Now I'm going to use the difference of squares pattern, {{{a^2-b^2=(a+b)(a-b)}}} to factor the right side, You might wonder: "But 3 is not a perfect square!?" And you'd be right. It's not the square of a whole number. But <i>every</i> positive number is the square of something. In this case 3 is the square of {{{sqrt(3)}}}. So we can factor the right side with this pattern:
{{{0 < (x + sqrt(3))(x-sqrt(3))}}}<br>
The reasons we got a zero and then factored are that the inequality above tells us something important. It tells us that a product (the right side) is greater than zero. A synonym for "greater than zero" is positive. So this tells us that a product of two numbers is positive. You have probably known for a long time that to get a positive result when multiplying two numbers you must have:<ul><li>Two positive numbers; or</li><li>Two negative numbers</li></ul>So our solution to this problem will be "all the x values that make both factors positive or both factors negative." All we have to do is express this logic in the form a inequalities we can solve.<br>
For "both factors are positive" I hope the following makes sense:
(x+sqrt(3) > 0 and x-sqrt(3)>0)
For "both factors are negative" I hope the following makes sense:
(x+sqrt(3) < 0 and x-sqrt(3)<0)
For "both factors positive or both factors negative" I hope that the following makes sense:
(x+sqrt(3) > 0 and x-sqrt(3)>0) or (x+sqrt(3) < 0 and x-sqrt(3)<0)
Now we just solve these:
(x > -sqrt(3) and x>sqrt(3)) or (x < -sqrt(3) and x<sqrt(3))<br>
The first pair both say that x is greater than something. With the "and" between them we are interested only in x's that are greater than both. I hope it makes sense that if x is greater than the larger number then it would automatically be greater than the smaller number, too. So {{{x > sqrt(3)}}} is all we need.<br>
The second pair both say that x is less than something. With the "and" between them we are interested only in x's that are less than both. I hope it makes sense that if x is less than the smaller number then it would automatically be less than the larger number, too. So {{{x < -sqrt(3)}}} is all we need.<br>
So our simplified solution is:
(x>sqrt(3)) or (x < -sqrt(3))