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 $|a+2|>2$.
\($x_1,x_2$\)
\($|a+2|>2$\)
\($\frac 1{x_1}+\frac 1{x_2}<1$\)
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