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Question:

The number of negative integral values of $m$ for which the expression $x^2+2(m-1)x+m+5$ is positive $\forall$ $x>1$ is?

For me, solving this question if the parameter "$\forall$ $x>1$" was not given would be quite easy. But how do I solve it under the given parameter?

General Kenobi
  • 439

3 Answers3

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There are two ways for having $\;p(x)=x^2+2(m-1)x+m+5>0\enspace\forall x>1$:

  • either $p(x)$ has no real root, which means the reduced discriminant $\Delta'=(m-1)^2-(m+5)=(m+1)(m-4)<0$ is negative;
  • or $p(x)$ has real roots, say $\xi_0\le \xi_1$ , i.e. $\Delta'\ge 0$. In this case, the condition means $1$ is on the right of the interval of the roots. This is satisfied if and only if $p(1)\ge 0$ ($1$ is outside of the interval of the roots or is a root) and, using Vieta's relations, $1>\frac12(\xi_0+\xi_1)=1-m$.
Bernard
  • 175,478
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  1. $f(1) >0$
  2. $\displaystyle -\frac{b}{2a}<0$

Symbols have their usual meaning

SarGe
  • 3,010
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Regarding this question geometrically, the equation for the described parabola in "vertex form" is $$ y \ \ = \ \ x^2 \ + \ 2(m-1)x \ + \ (m+5) \ \ = \ \ \left[ \ x \ + \ (m - 1) \ \right]^2 \ + \ (m+5) \ - \ (m-1)^2 \ \ $$ $$ = \ \ \left[ \ x \ - \ (1 - m) \ \right]^2 \ - \ (m^2 - 3m - 4) \ \ = \ \ \left[ \ x \ - \ (1 - m) \ \right]^2 \ - \ [ \ (m + 1)·(m - 4) \ ] \ \ . $$ We also note that the $ \ y-$intercept for the parabola is at $ \ y \ = \ m+5 \ \ . $

Hence, the vertex of this "upward-opening" parabola is "to the left" of the $ \ y-$axis for $ \ 1 - m < 0 \ \Rightarrow \ m > 1 \ \ . $ The $ \ y-$intercept is $ \ y > 5+1 = 6 \ \ , $ so the function is positive for $ \ x > 0 \ \ . $ For $ \ m = 1 \ \ , $ the equation is $ \ x^2 + 6 \ \ . $

These statements are prelude to addressing the behavior of the parabola when the vertex is "to the right" of the $y-$axis. The $ \ y-$coordinate of the vertex is positive for $ \ (m + 1)·(m - 4) \ < \ 0 \ \Rightarrow \ -1 < m < 4 \ \ $ ; the vertex becomes an $ \ x-$intercept for $ \ m = -1 \ \ . $ As the $ \ y-$intercept of the parabola is positive for $ \ -5 < m < 0 \ \ , $ the parabola is "on or above" the $ \ x-$axis for $ \ m = -1 \ \ $ [its equation is $ \ y \ = \ (x-2)^2 \ \ ] \ . $ For $ \ m < -1 \ \ , $ the vertex is "below" the $ \ x-$axis and its $ \ x-$coordinate is $ \ x > 2 \ \ , $ so the function is not positive for all $ \ x > 1 \ \ . $

So there is just one negative integral value for $ \ m \ \ ( m = -1 ) \ $ for which the quadratic function is non-negative for $ \ x > 1 \ \ , $ but none at all for which it is exclusively positive.