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I want to solve the inequality $2(x- \sqrt{2x^{2}-3x-5})\geq 1$.

First I find where the root square is well defined, that is, where $2x^{2}-3x-5 = (x+1)(2x-5) \geq 0.$

Doing a table of sings I get $\sqrt{2x^{2}-3x-5}$ is well defined in $(-\infty, -1] \cup [\frac{5}{2}, \infty)$.

Now, solving the inequality $2(x- \sqrt{2x^{2}-3x-5})\geq 1$, I do the following

$2x-2\sqrt{2x^{2}-3x-5}\geq 1$

$2x-1 \geq 2\sqrt{2x^{2}-3x-5}$

then

$(2x-1)^{2} \geq 4(2x^{2}-3x-5)$

then

$0 \geq 4x^{2}-8x-21 = (2x-7)(2x+3)$

doing a table of sings I get that the solutions for this inequality is $[-\frac{3}{2},\frac{7}{2}]$.

Finally, intersecting with the set where $\sqrt{2x^{2}-3x-5}$ is well defined I get the solution set is $[-\frac{3}{2},-1] \cup [\frac{5}{2},\frac{7}{2}]$.

But when I plot the function $2x-2\sqrt{2x^{2}-3x-5}- 1$ I get that the solution is $[\frac{5}{2},\frac{7}{2}]$.

enter image description here

What am I doing wrong?

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  • $\begingroup$ Squaring it has introduced an extraneous answer. $\endgroup$ – Ian Miller Feb 25 '16 at 7:45
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When you have $$2x-1\ge 2\sqrt{2x^2-3x-5}\ \ (\ge 0)$$ you have to have $$2x-1\ge 0.$$

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