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 December 24th, 2014, 01:32 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, $\displaystyle x^2+2ax+2a^2+a+1=0$ $\displaystyle b^2-4ac< 0$ $\displaystyle 4a^2-8a^2-4a-8< 0$ $\displaystyle a^2+a+2> 0$ I'm stuck at here. How should I continue?
 December 24th, 2014, 01:40 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. Thanks from jiasyuen
December 24th, 2014, 11:00 AM   #3
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Hello, jiasyuen!

Quote:
 Prove that, for all real values of $a$, the equation $x^2+2ax+2a^2+a+1\:=\:0$ has no real roots for $x$. My attempts: $x^2+2ax+2a^2+a+1=0$ $b^2-4ac< 0$ $4a^2-8a^2-4a-8< 0$ $a^2+a+2> 0$ I'm stuck at here. How should I continue?

You have a parabola: $\:y \:=\:x^2 + x+2$
When is it positive (above the x-axis)?

It is an up-opening 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 x-axis.
Therefore, the discriminant is always negative.

December 24th, 2014, 02:37 PM   #4
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Quote:
 Originally Posted by jiasyuen 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, $\displaystyle x^2+2ax+2a^2+a+1=0$ $\displaystyle b^2-4ac< 0$ $\displaystyle 4a^2-8a^2-4a-8< 0$ $\displaystyle a^2+a+2> 0$ I'm stuck at here. How should I continue?
$\displaystyle x^2+2ax+2a^2+a+1=(x+a)^2+a^2+a+1$
$\displaystyle a^2+a+1>0$ if a is real.
$\displaystyle (x+a)^2 \ge 0$ if x and a are real.
Therefore no real roots.

December 24th, 2014, 08:02 PM   #5
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Quote:
 Originally Posted by jiasyuen 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^2-4ac< 0 \ \ \ \$<---- This "a" variable is different from the "a" variable in the given equation, so you should use different letters. ** $\displaystyle 4a^2-8a^2-4a-8< 0 \ \ \ \ \$incorrect $\displaystyle a^2+a+2> 0 \ \ \ \ \$incorrect I'm stuck at here. How should I continue?
**$\displaystyle \ \ B^2 - 4AC < 0$

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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