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Im looking for an efficent method of solving the following inequality: $$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 <0$$

I've tried first determining when the absolute value will be positive or negative etc, and than giving it the signing in accordance to range it is in, bur it turned out to be quite complex and apperently also wrong. Are there any other ways?

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4 Answers 4

up vote 2 down vote accepted

When $t \equiv \frac{x-3}{x+1} > 0$, the solution to $ t^2 - 7t + 10 <0$ is $ 2<t<5$. (Factor the quadratic.)

When $t < 0$, the solution to $ t^2 + 7t + 10 <0$ is $ -2<t<-5$.

On the $t < 0$ branch, we need to solve $-2 < \frac{x-3}{x+1} < 5$. We break this up into two possibilities, $x < -1$ and $x > -1$, because when we multiply through by $x+1$ in the $x < -1$case we have to flip the sense of the the inequality.

When $x < -1$ we then get on one side $-2(x+1) < x-3$, which gives $x > +\frac{1}{3}$ and this does not work. But for $x > -1$ we get $$\begin{array}{c} -5(x+1) < x-3 < 2(x-1) \\ 6x > -2 \rightarrow x > -\frac{1}{3} \\ 3x < +1 \rightarrow x < +\frac{1}{3} \end{array} $$ which has solution $$ -\frac{1}{3} < x < \frac{1}{3} $$

On the $t > 0$ branch, we need to solve $2 < \frac{x-3}{x+1} < 5$. We again break this up into two possibilities, $x < -1$ and $x > -1$.

When $x < -1$ we then get on one side $2(x+1) > x-3$, which gives $x > -2$, and $5(x+1) < x-3$, which gives $x > -5$. For $x > -1$ we get $2(x+1) < x-3$, which holds if $x < -5$, which contradicts $x > -1$ so that case does not give any solutions.

Thus the answer combines the two solution regions:

$$ \left( -5 < x < 2 \right) \bigcup \left( -\frac{1}{3} < x < \frac{1}{3} \right) $$

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very detalied, thanks –  Bak1139 Jul 31 at 18:11

Substituting $t=|\frac {x-3}{x+1}|$ will help. Can you see it?

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This is a better substitution than the one suggested by @Elimination, I think –  MPW Jul 7 at 17:07
    
@MPW, that is because we have the advantage that $(|x|)^{2}=x^{2}$, –  Indrayudh Roy Jul 7 at 17:09
1  
Well, I was thinking that your substitution gives a simple quadratic inequality "$t^2-7t+10<0$", whereas the other substitution gives something about as difficult as the original problem, "$t^2-7|t| + 10 < 0$". –  MPW Jul 7 at 17:14
1  
Also, I should add that we must hope OP is only considering real solutions. Otherwise, we are in trouble, aren't we -- we will have to tread much more carefully in the complex domain since we lose the equality you mentioned (that is, $|z|^2$ need not equal $z^2$ for general $z\in\mathbb C$) –  MPW Jul 7 at 17:17
    
@MPW, the question is meaningless in the complex domain since order is meaningless in complex numbers. –  Indrayudh Roy Jul 7 at 17:21

Hint (for a start): $$t = \frac{x-3}{x+1}$$

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$$\left(\frac{x-3}{x+1}\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \\ \left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 $$

$$\Delta=\left(-7\right)^2-4 \cdot 10=49-40=9$$

$$\left|\frac{x-3}{x+1}\right|_{1,2}=\frac{-\left(-7\right) \pm \sqrt{\Delta}}{2}=\frac{7 \pm \sqrt{9}}{2}=\frac{7 \pm 3}{2}$$

$\left(\left|\frac{x-3}{x+1}\right|\right)^2-7 \left|\frac{x-3}{x+1}\right|+ 10 \lt 0 \Rightarrow \left|\frac{x-3}{x+1}\right| \in \left(\frac{7-3}{2},\frac{7+3}{2}\right)=\left(2,5\right)$

$$\left|\frac{x-3}{x+1}\right|\gt 2 \Rightarrow \frac{x-3}{x+1}\lt -2 \text{ OR } \frac{x-3}{x+1}\gt 2 $$ and $$\left|\frac{x-3}{x+1}\right|\lt 5 \Rightarrow -5\lt \frac{x-3}{x+1}\lt 5 $$

Solve at each case for $x$.

Can you continue?

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