+0  
 
0
90
1
avatar

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 $|a+2|>2$.

Guest Jan 12, 2018
Sort: 

1+0 Answers

 #1
avatar+82937 
+1

 

\($x_1,x_2$\)

 

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

 

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

 

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

CPhill  Jan 15, 2018
edited by CPhill  Jan 15, 2018

6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details