For the roots to be rational, the discriminant must be = 0, or a positive PERFECT SQUARE integer, such as 1, or 4, or 9, etc.

When the discriminant is set as ≥ 0, the result is the following quadratic inequality: 

Solving for p gives 2 IMAGINARY numbers. 

RELATED QUESTIONS

Show that the roots of the equation (x-p)(x-q)=2 are real and distinct for all real... (answered by Edwin McCravy,ikleyn)

For this problem: All the roots of
x^2 + px + q = 0 are real, where p and q are real... (answered by Edwin McCravy,AnlytcPhil,Plocharczyk,mccravyedwin)

All the roots of
x^2 + px + q = 0 are real, where p and q are real numbers. Prove that... (answered by ikleyn)

For this problem: All the roots of
x^2 + px + q = 0 are real, where p and q are real... (answered by Edwin McCravy)

So for this question(All the roots of
x^2 + px + q = 0 are real, where p and q are real... (answered by ikleyn)

given that the equation Z^2 + (p+iq)Z + r + is = 0 where p,q,r,s are real and non-zero... (answered by richard1234)

please help me : 

given that for all values x:
3x²+12x+5= p(x+q)²+r
 
a) find the... (answered by stanbon)

If p and q are the roots of the equation 2x^2-x-4=0. Find the equation whose roots are... (answered by Fombitz,youngguru23)

If p and q are the roots of the equation 
{{{2x^2-x-4=0}}} 
Find the equation whose... (answered by ikleyn)