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This is the question I have in front of me. Clearly, I need to find the range of values of $x$ for the given inequality.

Having taken a cue from a similar question, here's how I approached it -

Let $ a = x^2 + 3x + 1$ $\implies ax^2 + 3ax - (3a + 5) \geq 0$

Now, for this quadratic expression to be greater than or equal to $0$, the coefficient of $x^2$ must be $>0$, with the discriminant being equal to $0$. But, the coefficient of $x^2$ i.e. $a = x^2 + 3x + 1$ can take negative values also, which does not ensure the requirement.

So, how shall I go about this one? Any help will be appreciated. Thanks.

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  • $\begingroup$ just a note I had to use inspect in my browser just to get into this question $\endgroup$ – user451844 Aug 26 '17 at 16:40
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    $\begingroup$ $$\begin{array}{l}\text{Is this a question which I see before me,}\cr \text{The handle toward my hand? Come, let me clutch thee.}\cr \text{I have thee not, and yet I see thee still.}\cr \text{Art thou not, fatal vision, sensible}\cr \text{To feeling as to sight? Or art thou but}\cr \text{A question of the mind, a false creation,}\cr \text{Proceeding from the heat-oppressèd brain?}\end{array}$$ $\endgroup$ – Will Jagy Aug 26 '17 at 16:45
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not quite. Let $$ b = x^2 + 3 x - 1 $$ $$ (b+2)(b - 2) \geq 5 $$ $$ b^2 \geq 9 $$ $$ b \leq -3 \; \; \mbox{OR} \; \; b \geq 3 $$

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  • $\begingroup$ Well, the answer comes out to be a null set. Thanks @WillJagy. That was of great help. $\endgroup$ – Saksham Aug 26 '17 at 17:23
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    $\begingroup$ @Saksham it is not a null set. Get some graph paper and carefully graph $y = x^2 + 3x + 2$ and then $y = x^2 +3x-4$ ... printablepaper.net/category/graph $\endgroup$ – Will Jagy Aug 26 '17 at 17:36
  • $\begingroup$ Yeah I realised that soon after writing the comment, but forgot to make the amends here. $\endgroup$ – Saksham Aug 27 '17 at 5:24
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hint

Observe that if

$$f (x)=(x^2+3x+1)(x^2+3x-3) $$

then $$f (1)=f (-1)=f (-2)=f (-4)=5$$ thus

$$f (x)-5=$$ $$(x-1)(x+1)(x+2)(x+4) $$

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    $\begingroup$ This solution is really elegant, even more than the other solution. +1 $\endgroup$ – John Lou Aug 26 '17 at 17:38
  • $\begingroup$ @JohnLou Thanks a lot for this.:). $\endgroup$ – hamam_Abdallah Aug 26 '17 at 18:10

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