🚀 go-pugleaf

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Peter Brewer wrote:

> Consider this inequality:    x^2 +2x -3 > 0
> Which factorises to:    (x+3)(x-1) > 0
> The natural thing to say is therefore     x > -3 and x > 1
> However, if this inequality is solved graphically then the solution is
> x < -3 and x > 1
> which I believe is the correct answer. Can anyone tell me why the first
> technique is valid?
>
>     Thanks in advance,
>             Pete

 Hello Pete,

It is very simple .. and natural:

Suppose I give you the inequality A * B > 0, where A and B are some numbers
(positive or negative).

Q. How is it possible to get a positve product when multiplying these two
numbers A and B? (What are the circumstances or cases?)

A. CASE  I: If A is positive (A > 0) AND B is positive (B > 0), then A * B
> 0.

   OR

   CASE II: If A is negative (A < 0) AND B is negative (B < 0), then A * B
> 0.

These are the only cases to consider.  There are, however, two more cases
which will make A * B < 0. And there are other cases that will make A * B =
0, but we don't need them for this question.

Let A = x + 3
Let B = x - 1

In CASE  I, we get (x + 3 > 0) AND (x - 1 > 0)
Well x + 3 > 0 is the same as x > -3
And  x - 1 > 0 is the same as x > 1

Therefore, if (x > -3) AND (x > 1), then (x + 3)(x - 1) > 0.

Q. What values of x satisfy the requirement that (x > -3) AND (x > 1)
A. For example, if you choose x = -2, then (-2 > -3) (Check), but is (-2 >
1)? NO!

As long as (x > 1), then this will meet BOTH requirements.  All we have
done is condense the statement "(x > -3) AND (x > 1)" to the new statement
"(x > 1)".  They mean the same.

Therefore, if (x > 1), then (x + 3)(x - 1) > 0

You are now done with CASE I.

Q. What other values of x will make (x + 3)(x - 1) > 0
A. Just take a look at the other cases (CASE II) and following the same
logic we applied to CASE I above.

You will end up with the statement:

if (x < -3), then (x + 3)(x - 1) > 0


That is why x > 1 and x < -3 are the solution sets to the inequality (x +
3)(x - 1) > 0


I hope this helps.. BUT I strongly recommend that you ask your teacher for
help.  You have a right to ask them and they are ( or at least should be )
more than willing to help you.


El-Noor


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<HTML>


<P>Peter Brewer wrote:
<BLOCKQUOTE TYPE=CITE>Consider this inequality:&nbsp;&nbsp;&nbsp; x^2 +2x
-3 > 0
<BR>Which factorises to:&nbsp;&nbsp;&nbsp; (x+3)(x-1) > 0
<BR>The natural thing to say is therefore&nbsp;&nbsp;&nbsp;&nbsp; x > -3
and x > 1
<BR>However, if this inequality is solved graphically then the solution
is&nbsp;&nbsp;&nbsp; x &lt; -3 and x > 1
<BR>which I believe is the correct answer. Can anyone tell me why the first
technique is valid?

<P>&nbsp;&nbsp;&nbsp; Thanks in advance,
<BR>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Pete</BLOCKQUOTE>
<FONT SIZE=-1>&nbsp;</FONT><TT>Hello Pete,</TT><TT></TT>

<P><TT>It is very simple .. and natural:</TT><TT></TT>

<P><TT>Suppose I give you the inequality A * B > 0, where A and B are some
numbers (positive or negative).</TT><TT></TT>

<P><TT>Q. How is it possible to get a positve product when multiplying
these two numbers A and B? (What are the circumstances or cases?)</TT><TT></TT>

<P><TT>A. CASE&nbsp; I: If A is positive (A > 0) AND B is positive (B >
0), then A * B > 0.</TT><TT></TT>

<P><TT>&nbsp;&nbsp; OR</TT><TT></TT>

<P><TT>&nbsp;&nbsp; CASE II: If A is negative (A &lt; 0) AND B is negative
(B &lt; 0), then A * B > 0.</TT><TT></TT>

<P><TT>These are the only cases to consider.&nbsp; There are, however,
two more cases which will make A * B &lt; 0. And there are other cases
that will make A * B = 0, but we don't need them for this question.</TT><TT></TT>

<P><TT>Let A = x + 3</TT>
<BR><TT>Let B = x - 1</TT><TT></TT>

<P><B><TT>In CASE&nbsp; I</TT></B><TT>, we get (x + 3 > 0) AND (x - 1 >
0)</TT>
<BR><TT>Well x + 3 > 0 is the same as x > -3</TT>
<BR><TT>And&nbsp; x - 1 > 0 is the same as x > 1</TT><TT></TT>

<P><TT>Therefore, if (x > -3) AND (x > 1), then (x + 3)(x - 1) > 0.</TT><TT></TT>

<P><TT>Q. What values of x satisfy the requirement that (x > -3) AND (x
> 1)</TT>
<BR><TT>A. For example, if you choose x = -2, then (-2 > -3) (Check), but
is (-2 > 1)? NO!</TT><TT></TT>

<P><TT>As long as (x > 1), then this will meet BOTH requirements.&nbsp;
All we have done is condense the statement "(x > -3) AND (x > 1)" to the
new statement "(x > 1)".&nbsp; They mean the same.</TT><TT></TT>

<P><TT>Therefore, if (x > 1), then (x + 3)(x - 1) > 0</TT><TT></TT>

<P><TT>You are now done with CASE I.</TT><TT></TT>

<P><TT>Q. What other values of x will make (x + 3)(x - 1) > 0</TT>
<BR><TT>A. Just take a look at the other cases (CASE II) and following
the same logic we applied to CASE I above.</TT><TT></TT>

<P><TT>You will end up with the statement:</TT><TT></TT>

<P><TT>if (x &lt; -3), then (x + 3)(x - 1) > 0</TT>
<BR><TT></TT>&nbsp;<TT></TT>

<P><TT>That is why x > 1 and x &lt; -3 are the solution sets to the inequality
(x + 3)(x - 1) > 0</TT>
<BR><TT></TT>&nbsp;<TT></TT>

<P><TT>I hope this helps.. BUT I strongly recommend that you ask your teacher
for help.&nbsp; You have a right to ask them and they are ( or at least
should be ) more than willing to help you.</TT>
<BR><TT></TT>&nbsp;<TT></TT>

<P><TT>El-Noor</TT>
<BR><TT></TT>&nbsp;</HTML>

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