More precise way of solving inequalityI need to solve this function: | x 2 − 1 | ≥ 2 x − 2 I solved this equation:For x &lt; 0, the solution is non existing, here I got negative root, when I tried to solve quadratic function and for x ≥ 0 I got points x 1 = − 1 and x 2 = 3.My question is:How do I set the solution of equation. Is there any procedure, with wich I can determine is equation valid for [ − ∞ , − 1 ] ∨ [ 3 , + ∞ ].I know that I can just set the numbers and see the result, but I just want to now is there any other different way to do this.

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More precise way of solving inequalityI need to solve this function:  |      x    2    −  1  |  ≥  2  x  −  2  I solved this equation:For   x  &amp;lt;  0, the solution is non existing, here I got negative root, when I tried to solve quadratic function and for   x  ≥  0 I got points       x    1    =  −  1 and       x    2    =  3.My question is:How do I set the solution of equation. Is there any procedure, with wich I can determine is equation valid for   [  −  ∞  ,  −  1  ]  ∨  [  3  ,  +  ∞  ].I know that I can just set the numbers and see the result, but I just want to now is there any other different way to do this.

Omari Lane · Accepted Answer

Step 1You have two cases :Case 1 : When       x    2    −  1  ≥  0, you have       x    2    −  1  ≥  2  x  −  2..Case 2 : When       x    2    −  1  &amp;lt;  0, you have   −  (      x    2    −  1  )  ≥  2  x  −  2..Step 2The answer is  &quot;      x    2    −  1  ≥  0     and         x    2    −  1  ≥  2  x  −  2  &quot;     or     &quot;      x    2    −  1  &amp;lt;  0     and     −  (      x    2    −  1  )  ≥  2  x  −  2.  &quot;

Leonard Dyer · Accepted Answer

Step 1As             |        x          |        =          {                                    x                                              if                         x            ≥            0                                                −            x                                              if                         x            &amp;lt;            0                                    If       x    2    −  1  ≥  0    ⟺    x  ≥  1 or   x  ≤  −  1  ,we get      x    2    −  1  ≥  2  x  −  2    ⟺        x    2    −  2  x  +  1  ≥  0    ⟺    (  x  −  1      )    2    ≥  0which is trueStep 2If       x    2    &amp;lt;  1    ⟺    −  1  &amp;lt;  x  &amp;lt;  1              (  1  )  ,we get  −  (      x    2    −  1  )  &amp;gt;  2  x  −  2    ⟺        x    2    +  2  x  −  3  &amp;lt;  0    ⟺    (  x  +  3  )  (  x  −  1  )  &amp;lt;  0    ⟺    −  3  &amp;lt;  x  &amp;lt;  1              (  2  )Now using   (  1  )  ,  (  2  )  −  1  &amp;lt;  x  &amp;lt;  1So, x can assume any real value

More precise way of solving inequality. |x^2-1| >= 2x-2

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