Find the value(s) or range of values of p for which the equation x^2+px+8=p has a positive root and a negative root.

Question

Ans= p>8

Answer 1

x² + px + 8 = p must have both a positive and a negative real root.

 

This means x² + px + (8-p) = 0 satisfies this.

 

Using the quadratic formula, the roots are x = ( -p ± √(p² - 4(8-p) )  )/2

 

So there is both a positive and negative root if ( note that under the radical we have p2 +4p -32)

p² +4p - 32 > p^2, or 4p - 32 > 0, or p > 8.   If p < 8 then the + value of the radical is not sufficient to overcome the negative value of -p.  

 

So the solution is p > 8.