SOLUTION: show that the quadratic equation x^2+2px＋p(p－1)＝0 has two unequal real roots for any positive values of p.

Algebra ->  Equations -> SOLUTION: show that the quadratic equation x^2+2px＋p(p－1)＝0 has two unequal real roots for any positive values of p.      Log On


   

Question 993998: show that the quadratic equation x^2+2px＋p(p－1)＝0 has two unequal real roots for any positive values of p.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2%2B2px%2Bp%5E2-p=0

x=%28-2p%2B-+sqrt%28%282p%29%5E2-4%28p%5E2-p%29%29%29%2F%282%29

x=%28-2p%2B-+2%2Asqrt%28p%5E2-%28p%5E2-p%29%29%29%2F%282%29

x=%28-p%2B-+sqrt%28p%5E2-p%5E2%2Bp%29%29%2F1

highlight%28x=-p%2B-+sqrt%28p%29%29
and p is given to be a real number p%3E0. This means the square root expression does not disappear, and the plus&minus nature of the two roots insures that x is two unequal values.

x=-p-sqrt%28p%29 or x=-p%2Bsqrt%28p%29