Find the value of p for which the quadratic equation px²+6x+1=0 has real roots
Solution:
Given equation, px²+6x+1=0Here, a = p, b = 6, c = 1Since roots are real,So, b² - 4ac ≥ 0(p)² - 4 x 6 x 1 ≥ 0p² – 24 ≥ 0p² ≥ 24p ≥ ±√24p ≥ ±2√6The value of p so that the equation px²+6x+1=0 has real roots is p ≥ 2√6 or p ≥ -2√6.