Quadratic inequalities - why does the solution seem different for ( x − a ) ( x − b ) &lt; 0 and ( x − a ) ( x − b ) &gt; 0 3 2 x − 10 ⋅ 3 x + 9 &gt; 0 ( 3 x − 9 ) ( 3 x − 1 ) &gt; 0 3 x ∈ ( − ∞ ; 1 ) ∪ ( 9 ; + ∞ )So to find x we need to solve: 3 x &gt; 9 and 3 x &gt; 1 (1)and we get: x ∈ ( − ∞ ; 0 ) ∪ ( 2 ; + ∞ )Second example: 5 2 x − 6 ⋅ 5 x + 5 &lt; 0 ( 5 x − 5 ) ( 5 x − 1 ) &lt; 0 5 x ∈ ( 1 ; 5 )So to find x we need to solve: { 5 x &lt; 5 5 x &gt; 1 (2)and we get: x ∈ ( 0 ; 1 )My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)

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Quadratic inequalities - why does the solution seem different for   (  x  −  a  )  (  x  −  b  )  &amp;lt;  0 and   (  x  −  a  )  (  x  −  b  )  &amp;gt;  0      3          2      x        −  10  ⋅      3    x    +  9  &amp;gt;  0  (      3    x    −  9  )  (      3    x    −  1  )  &amp;gt;  0      3    x    ∈  (  −  ∞  ;  1  )  ∪  (  9  ;  +  ∞  )So to find x we need to solve:      3    x    &amp;gt;  9 and       3    x    &amp;gt;  1 (1)and we get:  x  ∈  (  −  ∞  ;  0  )  ∪  (  2  ;  +  ∞  )Second example:      5          2      x        −  6  ⋅      5    x    +  5  &amp;lt;  0  (      5    x    −  5  )  (      5    x    −  1  )  &amp;lt;  0      5    x    ∈  (  1  ;  5  )So to find x we need to solve:      {                                        5            x                    &amp;lt;          5                                                  5            x                    &amp;gt;          1                         (2)and we get:  x  ∈  (  0  ;  1  )My question is why in first example I get answer by solving 2 equations (1) seperately but in second example I get answer by solving system of equations (2)

laure6237ma · Accepted Answer

When you have   a  b  &amp;gt;  0, you can have   a  &amp;gt;  0  ,  b  &amp;gt;  0,  a  &amp;gt;  0  ,  b  &amp;gt;  0 or   a  &amp;lt;  0  ,  b  &amp;lt;  0,  a  &amp;lt;  0  ,  b  &amp;lt;  0. So for your first example you should have   [      3    x    &amp;gt;  9 and       3    x    &amp;gt;  1  ] or   [      3    x    &amp;lt;  9 and       3    x    &amp;lt;  1  ]. You can combine each of the square brackets because one of the inequalities is always true when the other one is to get       3    x    &amp;gt;  9 or       3    x    &amp;lt;  1 (note the or, not and), leading to the solution you have. For the second, if   a  b  &amp;lt;  0 one of   a and   b is greater than zero and the other is less. So you can have   [      5    x    &amp;gt;  5 and       5    x    &amp;lt;  1  ] or   [      5    x    &amp;lt;  5 and       5    x    &amp;gt;  1  ]. In this case the inequalities in the first bracket are inconsistent, so that one can be ignored and we get the inequalities you cite.

Oakey1w · Accepted Answer

There is no difference: in the first case you should solve  (      3    x    &amp;gt;  9         and           3    x    &amp;gt;  1  )         or       (      3    x    &amp;lt;  9         and           3    x    &amp;lt;  1  )  ,and in the second case you should solve  (      5    x    &amp;gt;  5         and           5    x    &amp;lt;  1  )         or       (      5    x    &amp;lt;  5         and           5    x    &amp;gt;  1  )  .It just happens that   (      3    x    &amp;gt;  9 and       3    x    &amp;gt;  1  ) reduces to   (      3    x    &amp;gt;  9  ), that   (      3    x    &amp;lt;  9 and       3    x    &amp;lt;  1  ) reduces to   (      3    x    &amp;lt;  1  ), and that   (      5    x    &amp;gt;  5 and       5    x    &amp;lt;  1  ) is impossible.

Quadratic inequalities - why does the solution seem different for ( x −

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