Solve 1 x 2 − 1 ⩾ 1 x + 1

Question

Solve       1                  x        2            −      1        ⩾      1          x      +      1

rideonthebussp · Accepted Answer

Use the difference of two squares to get

\frac{1}{(x + 1) (x - 1)} \geq \frac{1}{x + 1}

Case 1: If

x + 1 > 0 \Rightarrow x > - 1

, we have:

\begin{matrix} (x \neq - 1, 1) & \frac{1}{x - 1} \geq 1 \end{matrix}

The

x \neq - 1, 1

comes from the denominator

(x + 1) (x - 1)

, which is undefined when

x = - 1, 1

.
If

x - 1 > 0 \Rightarrow x > 1

, then multiplying both sides by

x - 1

gives

1 \geq x - 1 \Rightarrow x \leq 2

. If

x - 1 < 0

, then because

x > - 1

, there are no solutions for this case. Hence the intersection of

x > - 1, x > 1

, and

x \leq 2

imply

1 < x \leq 2

.
Case 2: If

x + 1 < 0 \Rightarrow x < - 1

, we have:

\begin{matrix} (x \neq - 1, 1) & \frac{1}{x - 1} \leq 1 \end{matrix}

x - 1 > 0

is not possible in this case because

x < - 1

. Thus

x - 1 < 0

, or just

x < - 1

, and there are no more conditions.
So the solution to the inequality is

x < - 1

,

1 < x \leq 2

for

x \in R

.

Thomas Hubbard · Accepted Answer

We note on the side that       x    2    ≠  1 and multiply by       x    2    −  1, so need to distinguish two cases:      {                                        x            2                    −          1          &amp;gt;          0          →          1          ≥          x          −          1          ,                                                  x            2                    −          1          &amp;lt;          0          →          1          ≤          x          −          1.                        The second case is impossible and we are left with  x  &amp;lt;  −  1  ∨  1  &amp;lt;  x  ≤  2.You can also work in systematic way, using a table of sign variations,                                −        1                          1                          2                            x        +        1                    −                    0                    +                    +                    +                    +                    +                            x        −        1                    −                    −                    −                    0                    +                    +                    +                            2        −        x                    +                    +                    +                    +                    +                    0                    −                                  +                          |                    −                          |                    +                    0                    −

Solve 1 x 2 </msup> −

Answered question

Answer & Explanation

New Questions in Algebra I