1.

Solve: find the values of x​

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\large\underline{\sf{Solution-}}

GIVEN inequality is

\rm :\longmapsto\:( {x}^{2} + 3x + 1)( {x}^{2} + 3x - 3) \geqslant 5

Let we ASSUME that,

\red{\rm :\longmapsto\: {x}^{2} + 3x = y}

So, above inequality can be rewritten as

\rm :\longmapsto\:(y + 1)(y - 3) \geqslant 5

\rm :\longmapsto\: {y}^{2} - 3y + y - 3 \geqslant 5

\rm :\longmapsto\: {y}^{2} - 2y - 8 \geqslant 0

\rm :\longmapsto\: {y}^{2} - 4y + 2y - 8 \geqslant 0

\rm :\longmapsto\:y(y - <klux>4</klux>) + 2(y - 4) \geqslant 0

\rm :\longmapsto\:(y - 4)(y + 2) \geqslant 0

On substituting back the value of y, we get

\rm :\longmapsto\:( {x}^{2} + 3x - 4)( {x}^{2} + 3x + 2) \geqslant 0

\rm :\longmapsto\:\bigg( {x}^{2} + 4x - x - 4 \bigg) \bigg( {x}^{2} + 2x + x + 2 \bigg) \geqslant 0

\rm :\longmapsto\:\bigg(x(x + 4) - 1(x + 4) \bigg)\bigg(x(x + 2) + 1(x + 2) \bigg) \geqslant 0

\rm :\longmapsto\:(x + 4)(x - 1)(x + 2)(x + 1) \geqslant 0

So, breaking points are - 4, - 2, - 1, 1

Let check the interval for the sign.

\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf x \leqslant - 4 & \sf + \\ \\ \sf - 4 \leqslant x \leqslant - 2 & \sf - \\ \\ \sf - 2 \leqslant x \leqslant - 1 & \sf + \\ \\ \sf - 1 \leqslant x \leqslant 1 & \sf - \\ \\ \sf x \geqslant 1 & \sf + \end{array}} \\ \end{gathered}

So, we concluded that

\bf\implies \:x \in \: ( - \infty , - 4] \cup \: [ - 2, - 1] \cup \: [ 1, \infty )


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