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Let $a,b$ be real numbers, and let $x_1,x_2$ be the roots of the quadratic equation $x^2+ax+b=0$. Prove that if $x_1,x_2$ are real and nonzero, $\frac 1{x_1}+\frac 1{x_2}<1$, and $b>0$, then $|a+2|>2$.

 Jan 12, 2018
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\($x_1,x_2$\)

 

\($|a+2|>2$\)

 

\($\frac 1{x_1}+\frac 1{x_2}<1$\)

 

The sum of the roots =   -a  

And since  the  product of the roots  = b  >  0.....this implies that both roots have the same sign    

 

Let  x1  , x2   be the positive roots

So  -x1, -x2  are the negative roots

 

The  sum of the negative roots  results in

-x1 +   -x2  =   -a

- [ x1  + x2  ] =  -a      ⇒  x1  +  x2  = a

 

Therefore......a  in this  case   >  0  so

l a + 2  l  >  2    is satisfied

 

Now....if both positive roots   are   >  0   but  ≤ 2   ....then.....

 

1 /x1  +  1/x2   ≥ 1         but this violates the condition that  1/x1  + 1/x2  <  1

 

Therefore   x1, x2   must  be  >  2

 

Therefore

 

x1  + x2  =  -a

 

- [ x1  + x2  ]   =  a

 

But.....since   x1, x2  >  2   then their sum  must be  > 4

 

So....this implies that 

 

[x1 + x2]  >  4 

 

 - [ x1  + x2 ]  <  -4     ⇒    a  <  - 4

 

So.....in the case of two positive roots  .......

 

l a  + 2 l  >  2     is true

 

So......

 

l a  + 2  l  >  2   is proved for both positive and negative roots

 

 

cool cool cool

 Jan 15, 2018
edited by CPhill  Jan 15, 2018

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