quadratic equation

Question

If p,q,r are real numbers satisfying condition p+q+r=0 then roots of quadratic equation 3px^2+5qx+7r=0 are

Answer 1

     You can still use the quadratic formula even though the coefficients are variables. a= 3p, b=5q and c=7r. So,

x = - 5q +- sqrt( 25q^2 - 4(3p)(7r)) over 2(3p) 

x = [- 5q +- sqrt( 25q^2 - 84pr) ]/6p 

 

In order for this to be a real number, the discriminant must be >= 0. Otherwise you'd have sqrt of a negative which leads to complex roots. So, 

25q^2 - 84pr >= 0 and p+q+r = 0. solvig for r, r= -p-q and replacing, you'd have 

25q^2 - 84p( -p-q ) >= 0

25q^2 +84p^2 + 84pq >= 0       This does not have one particular answer, but rather an infinite number of p's and q's. For

                                                            example, if p and q are the same sign, the expression is always positive. So you may be looking

                                                            for a general solution?

 

So, x = - 5q +- sqrt( 25q^2 +84p^2 + 84pq )

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                                                           6p

 

Hope this helps a little bit.