Understanding Solutions to Quadratic Inequalities

[name_ask17]

Homework Statement

i know this seems basic, but read my attempt at solution.
x^2 -4>0

The Attempt at a Solution

i moved 4 over and took the square root of both sides to make the it x>(+)or(-)2
but that is not the answer, that is the error and we have to prove why it is the error. Any ideas here?

 

[LCKurtz]

Factor x²-4 and analyze the signs of the factors for various values of x. The sign of the product will be determined by the signs of the factors.

 

[phinds]

plug in x = -3

 

[Saitama]

Here what you can do is convert x²-4 to (x+2)(x-2).
So what you get is (x+2)(x-2)>0.
Therefore the domain comes out to be (-\infty,-2)\cup(2,\infty).

 

[HallsofIvy]

A product of two numbers is positive if and only if the two numbers have the same sign.

So x^2- 4= (x- 2)(x+ 2)&gt; 0 if and only if (1) x- 2> 0 and x+ 2> 0 or (2) x-2< 0 and x+2< 0.

x-2> 0 gives x> 2 and x+ 2> 0 gives x> -2. Both of those will be true if x> 2.

x-2< 0 gives x< 2 and x+ 2< 0 gives x< -2. Both of those will be true if x< -2.

That is where Panav-arora gets his answer of "(-\infty, -2)\cup (2, \infty)" (although it would have been better to give hints rather than simply the answer).

A more general way of solving complicated (non-linear) inequalities is to solve the associated equation. The equation associated with x^2- 4&gt; 0 is x^2- 4= 0 which has roots x= -2 and x= 2. Those two numbers divide the real number line into three intervals, x< -2, -2< x< 2, and x> 2. Since x^2- 4 is a continuous function, x= -2 and x= 2 are the only places where ">" can change to "<" and vice-versa. That is, we must have either ">" or "<" for all points on each interval and so it is enough to check for one point in each interval. If, for example, x= -3< -2, is x^2- 4&gt; 0? If yes, then x^2- 4&gt; 0 for all x< -2. 0 is between -2 and 2. If x= 0, is x^2- 4&gt; 0? If yes, then it is true for all x between -2 and 2. If false, then it is false for all numbers between -2 and 2. 3> 2. If x= 3, is x^2- 4&gt; 0? If it is true, then it is true for all x> 2.

 

[Saitama]

That is where Panav-arora gets his answer of "(-\infty, -2)\cup (2, \infty) (although it would have been better to give hints rather than simply the answer).

<br /> <br /> Ok I will try to provide hints <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> One easy way to find its domain is with the use of wavy-curve method which my teacher taught and its really useful.<br /> We got (x+2)(x-2)&gt;0<br /> Now if we set (x+2)(x-2)=0, we get two roots i.e. -2 &amp; 2. Plot these roots on the number. <br /> If we take the values greater than 2 and plugin it in (x+2)(x-2)&gt;0, we get a value greater than 0. Now we can directly know the intervals by the wavy curve method as i have shown in the diagram below:-<br /> <br /> <script class="js-extraPhrases" type="application/json"> { "lightbox_close": "Close", "lightbox_next": "Next", "lightbox_previous": "Previous", "lightbox_error": "The requested content cannot be loaded. Please try again later.", "lightbox_start_slideshow": "Start slideshow", "lightbox_stop_slideshow": "Stop slideshow", "lightbox_full_screen": "Full screen", "lightbox_thumbnails": "Thumbnails", "lightbox_download": "Download", "lightbox_share": "Share", "lightbox_zoom": "Zoom", "lightbox_new_window": "New window", "lightbox_toggle_sidebar": "Toggle sidebar" } </script> <div class="bbImageWrapper js-lbImage" title="8cc8615bb20f8ebcc4c392643bfd9bee6d2496ccb45bba1b4e6d38c9205327c82g.jpg" data-src="https://www.physicsforums.com/attachments/8cc8615bb20f8ebcc4c392643bfd9bee6d2496ccb45bba1b4e6d38c9205327c82g-jpg.143799/" data-lb-sidebar-href="" data-lb-caption-extra-html="" data-single-image="1"> <img src="https://www.physicsforums.com/attachments/8cc8615bb20f8ebcc4c392643bfd9bee6d2496ccb45bba1b4e6d38c9205327c82g-jpg.143799/" data-url="" class="bbImage" data-zoom-target="1" style="" alt="8cc8615bb20f8ebcc4c392643bfd9bee6d2496ccb45bba1b4e6d38c9205327c82g.jpg" title="8cc8615bb20f8ebcc4c392643bfd9bee6d2496ccb45bba1b4e6d38c9205327c82g.jpg" width="191" height="145" loading="lazy" decoding="async" /> </div><br /> <br /> In this case three intervals are formed as HallsofIvy stated i.e (1)x lies between -infinity to -2 (2)x lies between -2 to 2 (3)x lies between 2 to -infinity. So if we find what would be the value of (x+2)(x-2) in one of the interval i.e if it would be greater than zero or less than zero. &quot;+&quot; sign indicates that in this interval, the value of (x+2)(x-2) is greater than zero and &quot;-&quot; sign indicates that (x+2)(x-2) is less than zero. So after finding the sign of one interval we will switch the sign in the upcoming series or preceding series. So it would be easier to find the domain interval of the given inequality. If it was given to find the domain interval of (x+2)(x-2)&lt;0, then by looking at the diagram we could instantly say that the domain is (-2,2). This trick results to be very useful when you have a load of questions to solve or you are in shortage of time.<br /> Hope this was helpful! <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" />

 

[name_ask17]

So what you get is (x+2)(x-2)>0.

Wouldn't that still make the solution to this problem be x>plus or minus 2, becuase x>2 and x>-2?
My problem was that I have to find the solution to the problem, but the answer is not x>plus or minus 2 and I am confused as to why this is not the answer.

 

[phinds]

So answer me a question. Is zero greater that minus 2? Is zero inside the range of the answer to this problem?

All you had to do in the first place is just draw a rough graph of the function f(x) = x^2 - 4 and the solution is obvious. You don't need any fancy math AT ALL if you just want to get the right answer.

 

[name_ask17]

OHH! Okay, thank you for all your assistance. I got the answer. I was overthinking a basic problem. The number line method worked too. Thank you all again.

 

[Saitama]

Wouldn't that still make the solution to this problem be x>plus or minus 2, becuase x>2 and x>-2?
My problem was that I have to find the solution to the problem, but the answer is not x>plus or minus 2 and I am confused as to why this is not the answer.

What's the answer then?
And its not x>-2, it would be x<-2.