+0

# Still the same anon

0
288
3

This is the answer it is giving me for the equasion x^2+7x+53=11/3, compleatly diffrent from your answer, and it says it yealds {}:

Guest May 30, 2014

#3
+91465
+10

Hi anonymous,

Now I finally understand what your problem was. (Thank you Alan)  I am so glad that you have been persistent in your questioning.  This is what I want everyone to do!  If you do not understand our answer or you think we are wrong just keep at us until you are happy.  Remember though to try and work out WHY we don't understand your question or your problems.  That way we can help each other.

In this case higher mathematician, including me, forgot that most  student have never heard of imaginary numbers.  I am so glad that you have reminded me of this divide.

Thanks, I am genuinely grateful to you.

------------------------------------------------------------------

I just want to talk about it a little more:

x^2+7x+53=11/3

Think about this in this way (Alan has done it a bit different but visually this might be easier for you to understand.)

$$x^2+7x+53=\frac{11}{3}$$

I am going to split this up into two bits.   I'm going to let y = the left side and y=right side. Like this;

$$y=x^2+7x+53\mbox\qquad { and }\qquad y=\frac{11}{3}$$

Now I am going to graph these two functions.  Where they cross each other will be where  $$x^2+7x+53=\frac{11}{3}$$

NOTE:  This is how you always solve simultaneous equations when doing it graphically.

You can see from the graph that it is impossible for

$$y\quad \mbox{to equal}\quad x^2+7x+53\mbox\quad {and}\quad \frac{11}{3}\quad \mbox {both at the same time.}$$

So as Alan said, there are no real solutions, the answer in the set of REAL numbers is {} (an empty set.)

There are however complex number solutions - These are imaginary numbers! They are the answers that we were so determined to give you.  Sorry.

Melody  Jun 1, 2014
Sort:

#1
0

I don't know if you can see the picture, but it was a graph

Guest May 30, 2014
#2
+26402
+10

I can't see your graph, but here is one:

You can see that there is not going to be any solution in terms of real numbers; that's why you get {} - this means the empty set (i.e. nothing in it).

The only solutions exist in the complex number domain as the previous replies showed.

Alan  May 30, 2014
#3
+91465
+10

Hi anonymous,

Now I finally understand what your problem was. (Thank you Alan)  I am so glad that you have been persistent in your questioning.  This is what I want everyone to do!  If you do not understand our answer or you think we are wrong just keep at us until you are happy.  Remember though to try and work out WHY we don't understand your question or your problems.  That way we can help each other.

In this case higher mathematician, including me, forgot that most  student have never heard of imaginary numbers.  I am so glad that you have reminded me of this divide.

Thanks, I am genuinely grateful to you.

------------------------------------------------------------------

I just want to talk about it a little more:

x^2+7x+53=11/3

Think about this in this way (Alan has done it a bit different but visually this might be easier for you to understand.)

$$x^2+7x+53=\frac{11}{3}$$

I am going to split this up into two bits.   I'm going to let y = the left side and y=right side. Like this;

$$y=x^2+7x+53\mbox\qquad { and }\qquad y=\frac{11}{3}$$

Now I am going to graph these two functions.  Where they cross each other will be where  $$x^2+7x+53=\frac{11}{3}$$

NOTE:  This is how you always solve simultaneous equations when doing it graphically.

You can see from the graph that it is impossible for

$$y\quad \mbox{to equal}\quad x^2+7x+53\mbox\quad {and}\quad \frac{11}{3}\quad \mbox {both at the same time.}$$

So as Alan said, there are no real solutions, the answer in the set of REAL numbers is {} (an empty set.)

There are however complex number solutions - These are imaginary numbers! They are the answers that we were so determined to give you.  Sorry.

Melody  Jun 1, 2014

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