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#1
So I did this question from my textbook and to reach the answers, they multiply both sides by x (when x is already present on one side) to get a quadratic equation. My question is... why?

The way I did it, I just saw it as a linear equation and found x > 4/5. I know that they multiply both sides by x to get the SECOND inequality x <0 , but I just don't get the train of thought. What I'm trying to ask is, when I see a linear inequality, when do I have to multiply both sides by x to get the values for x? Is it in every case? Say for example, the one I just made up on the spot in the picture : 5x - 2 > 0. To find the values for this one, do I multiply both sides by x to make it a quadratic? If no... why?
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5 months ago
#2
If you look on the previous page of textbook in the pink "watch out" box it explains it.
but here is full explanation:

The reason you do this is because when you multiply by a negative you must change the direction of the inequality sign.
For example
-1/2 < 6 THIS IS TRUE
but let's say we multiply both sides by -2 to get:
1< -12 THIS IS OBVIOUSLY NOT TRUE
so we must change the sign when we multiply by a negative:
1 > -12 NOW THIS IS TRUE

In your question you have a variable, x, you don't know the whether x is postive or negative as it varies, so when you multiply it over do you change the sign or not? Well, you don't know, so to get around this problem we can force what we multiply over to be postive by squaring it (sqauring a negative makes it postive). We have not changed the equation if we multiply both sides by x^2 and we know x^2 is postive so we don't have to change the sign.

In short, when YOU FEEL YOU NEED TO MULTIPLY BY X, INSTEAD MULTIPLY BY X^2

Also for 5x-2> 0 you cannot make it
5x^2 - 2x > 0
because you have just commited the sin i warned you about: you multiplied by x.
Because when you multiply by x whether you flip by the inequality sign or not depends on the value of x.
If you were to plot that quadratic you would get x<0 or x>2/5
but obviously this is wrong coz if x was less than 0, for example x=-1 then 5(-1)-2 = -7 which is obviously not less than zero so doesn't satisfy original requirements.
Instead what you must do is multiply by x^2 to get
5x^3 -2x^2>0
and if you plot this you will see that the only time when the graph/y-value is above zero is when x>2/5 so this makes sense and matches the answer you would get if you just moved 2 over and divided by 5 to get x>2/5
(you can plot it on this website to see for yourself: https://www.desmos.com/calculator )

But of course for 5x-2 you don't need to go through all this trouble just do it normally.
Last edited by gyuigygh; 5 months ago
0
#3
(Original post by gyuigygh)
If you look on the previous page of textbook in the pink "watch out" box it explains it.
but here is full explanation:

The reason you do this is because when you multiply by a negative you must change the direction of the inequality sign.
For example
-1/2 < 6 THIS IS TRUE
but let's say we multiply both sides by -2 to get:
1< -12 THIS IS OBVIOUSLY NOT TRUE
so we must change the sign when we multiply by a negative:
1 > -12 NOW THIS IS TRUE

In your question you have a variable, x, you don't know the whether x is postive or negative as it varies, so when you multiply it over do you change the sign or not? Well, you don't know, so to get around this problem we can force what we multiply over to be postive by squaring it (sqauring a negative makes it postive). We have not changed the equation if we multiply both sides by x^2 and we know x^2 is postive so we don't have to change the sign.

In short, when YOU FEEL YOU NEED TO MULTIPLY BY X, INSTEAD MULTIPLY BY X^2

Also for 5x-2> 0 you cannot make it
5x^2 - 2x > 0
because you have just commited the sin i warned you about: you multiplied by x.
Because when you multiply by x whether you flip by the inequality sign or not depends on the value of x.
If you were to plot that quadratic you would get x<0 or x>2/5
but obviously this is wrong coz if x was less than 0, for example x=-1 then 5(-1)-2 = -7 which is obviously not less than zero so doesn't satisfy original requirements.
Instead what you must do is multiply by x^2 to get
5x^3 -2x^2>0
and if you plot this you will see that the only time when the graph/y-value is above zero is when x>2/5 so this makes sense and matches the answer you would get if you just moved 2 over and divided by 5 to get x>2/5
(you can plot it on this website to see for yourself: https://www.desmos.com/calculator )

But of course for 5x-2 you don't need to go through all this trouble just do it normally.
oh right I get it now thanks !
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