Solving an inequality

[happy21]

Please help me solve the following inequality:

What will be the exhaustive interval of real values of x such that [tex]\sqrt{12-4x}[/tex] > 1+ [tex]\sqrt{4x+4}[/tex]

Thx.

[nathi123]

[tex]\sqrt{12-4x}>1+\sqrt{4x+4}\Leftrightarrow\begin{array}{|l} 12-4x\ge0 \\ 4x+4\ge 0\\ 12-4x>1+2\sqrt{4x+4}+4x+4\end{array}[/tex]
[tex]\Leftrightarrow \begin{array}{|l}-1 \le x \le 3\\ 7-8x\ge0\\ 2\sqrt{4x+4} <7-8x\\\end{array}[/tex] [tex]\Leftrightarrow \begin{array}{|l} -1\le x \le \frac{7}{8}\\ 16x+16<49-112x+64x^{2} \end{array}[/tex].
[tex]\Leftrightarrow\begin{array}{|l} -1\le x \le\frac{7}{8}\\64x^{2} -128x + 33>0 \end{array}[/tex] [tex]\Leftrightarrow -1\le x<\frac{8-\sqrt{31}}{8}[/tex].

[Guest]

[tex]\sqrt{12-4x}>1+\sqrt{4x+4}\Leftrightarrow\begin{array}{|l} 12-4x\ge0 \\ 4x+4\ge 0\\ 12-4x>1+2\sqrt{4x+4}+4x+4\end{array}[/tex]
[tex]\Leftrightarrow \begin{array}{|l}-1 \le x \le 3\\ 7-8x\ge0\\ 2\sqrt{4x+4} <7-8x\\\end{array}[/tex] [tex]\Leftrightarrow \begin{array}{|l} -1\le x \le \frac{7}{8}\\ 16x+16<49-112x+64x^{2} \end{array}[/tex].
[tex]\Leftrightarrow\begin{array}{|l} -1\le x \le\frac{7}{8}\\64x^{2} -128x + 33>0 \end{array}[/tex] [tex]\Leftrightarrow -1\le x<\frac{8-\sqrt{31}}{8}[/tex].

Thx.

[happy21]

[tex]\sqrt{12-4x}>1+\sqrt{4x+4}\Leftrightarrow\begin{array}{|l} 12-4x\ge0 \\ 4x+4\ge 0\\ 12-4x>1+2\sqrt{4x+4}+4x+4\end{array}[/tex]
[tex]\Leftrightarrow \begin{array}{|l}-1 \le x \le 3\\ 7-8x\ge0\\ 2\sqrt{4x+4} <7-8x\\\end{array}[/tex] [tex]\Leftrightarrow \begin{array}{|l} -1\le x \le \frac{7}{8}\\ 16x+16<49-112x+64x^{2} \end{array}[/tex].
[tex]\Leftrightarrow\begin{array}{|l} -1\le x \le\frac{7}{8}\\64x^{2} -128x + 33>0 \end{array}[/tex] [tex]\Leftrightarrow -1\le x<\frac{8-\sqrt{31}}{8}[/tex].

Thx!