Question

Помогите, пожалуйста, решить подробно!
$|2x^{2} -3x-1|=x^{2} -3$

Answer 1

Решение и ответ:

$\displaystyle \left| {2{x^2}-3x-1} \right|={x^2}-3$

$\[\left\{ \begin{gathered}{x^2}-3 \geqslant 0 \hfill \\\left[\begin{gathered}2{x^2}-3x-1={x^2}-3 \hfill \\2{x^2}-3x-1=-{x^2}+3 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.;\left\{ \begin{gathered}{x^2}-3 \geqslant 0 \hfill \\\left[ \begin{gathered}2{x^2}-3x-1-{x^2}+3=0 \hfill \\2{x^2}-3x-1+{x^2}-3=0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.\]$

$\[\left\{\begin{gathered}{x^2}-3 \geqslant 0 \hfill \\\left[ \begin{gathered}{x^2}-3x+2=0 \hfill \\3{x^2}-3x-4=0 \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.;\left\{ \begin{gathered}x \geqslant \sqrt 3,\;x \leqslant -\sqrt 3 \hfill \\\left[ \begin{gathered}x=2,\;x=1 \hfill \\x=\frac{{3+\sqrt {57} }}{6},\;x=\frac{{3-\sqrt {57} }}{6} \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} \right.\]$

${x^2}-3\geqslant 0$

${x^2}\geqslant 3$

$x \geqslant \sqrt 3 ,\;x \leqslant -\sqrt 3$

${x^2}-3x+2=0$

$\displaystyle D={b^2}-4ac={(-3)^2}-4\cdot 1\cdot2=9-8=1$

$\displaystyle x=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{3\pm\sqrt 1}}{{2\cdot1}}=\frac{{3\pm1}}{2}$

$\displaystyle x=\frac{{3+1}}{2}=\frac{4}{2}=2$

$\displaystyle x=\frac{{3-1}}{2}=\frac{2}{2}=1$

$\displaystyle 3{x^2}-3x-4=0$

$\displaystyle D={b^2}-4ac={(-3)^2}-4\cdot3\cdot(-4)=9+48=57$

$\displaystyle x=\frac{{-b\pm\sqrt D}}{{2a}}=\frac{{3\pm\sqrt{57}}}{{2\cdot3}}=\frac{{3\pm\sqrt{57}}}{6}$

$\displaystyle x=\frac{{3+\sqrt{57}}}{6}\approx1.758$

$\displaystyle x=\frac{{3-\sqrt{57}}}{6}\approx-0.758$

Ответ: $\displaystyle x=2;\;\;x=\frac{{3+\sqrt {57} }}{6}$