December 24th, 2014, 12:32 AM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  Proving.
Prove that, for all real values of a, the equation $\displaystyle x^2+2ax+2a^2+a+1=0$ has no real roots for x. My attempts, $\displaystyle x^2+2ax+2a^2+a+1=0$ $\displaystyle b^24ac< 0$ $\displaystyle 4a^28a^24a8< 0$ $\displaystyle a^2+a+2> 0$ I'm stuck at here. How should I continue? 
December 24th, 2014, 12:40 AM  #2 
Senior Member Joined: Nov 2013 From: Baku Posts: 502 Thanks: 56 Math Focus: Geometry 
$\displaystyle \Delta = 4a^2  8a^2  4a  4 = 4(a^2 +a+1)$ $\displaystyle x_{1,2} = \frac{1}{2} \times (2a \pm \sqrt{4(a^2 +a+1)} \; ) = a \pm i \sqrt{a^2 +a+1}$. NOTE: $\displaystyle a^2 +a+1$ is always positive. 
December 24th, 2014, 10:00 AM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Hello, jiasyuen! Quote:
You have a parabola: $\:y \:=\:x^2 + x+2$ When is it positive (above the xaxis)? It is an upopening parabola. Where is its vertex? Vertex: $\:x \:=\: \frac{\text{}b}{2a} \:=\:\frac{\text{}1}{2(1)} \:=\:\text{}\frac{1}{2}$ Then: $\:y \:=\:\left(\text{}\frac{1}{2}\right)^2 + \left(\text{}\frac{1}{2}\right) + 2 \;=\;\frac{1}{4}  \frac{1}{2} + 2 \;=\;\frac{7}{4}$ The vertex is: $\:\left(\text{}\frac{1}{2},\,\color{red}{+}\frac{7}{4}\right)$ The parabola is always above the xaxis. Therefore, the discriminant is always negative.  
December 24th, 2014, 01:37 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,528 Thanks: 589  Quote:
$\displaystyle a^2+a+1>0 $ if a is real. $\displaystyle (x+a)^2 \ge 0$ if x and a are real. Therefore no real roots.  
December 24th, 2014, 07:02 PM  #5  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
A = 1 B = 2a C = $\displaystyle 2a^2 + a + 1$ $\displaystyle (2a)^2 \  \ 4(1)(2a^2 + a + 1) \ < \ 0 $ $\displaystyle 4a^2  8a^2  4a  4 \ < \ 0$ $\displaystyle 4a^2  4a  4 \ < \ 0 $ $\displaystyle 4(a^2 + a + 1) \ < \ 0 $ $\displaystyle a^2 + a + 1 \ > \ 0 $  

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