Find the values of p for which the equation 3x² - px + 5 = 0 has real roots.

For real roots,

Discriminant ≥ 0

or, b² - 4ac ≥ 0

The above equation is 3x² - px + 5 = 0
Comparing with ax² + bx + c = we obtain,
a = 3 , b = -p , c = 5

Putting values in b² - 4ac ≥ 0 we get,

$= (-p)^2 - 4 \times 3 \times 5 \ge 0 \\[1em] = p^2 - 60 \ge 0 \\[1em] \Rightarrow p^2 - (\sqrt{60})^2 \ge 0 \\[1em] \Rightarrow (p + \sqrt{60})(p -\sqrt{60}) \ge 0 \\[1em] \Rightarrow (p + 2\sqrt{15})(p - 2\sqrt{15}) \ge 0 \\[1em] \text{(Using) } (a + b)(a - b) \ge 0 = a \ge b \text{ or } a \le -b \text{ we get, } \\[1em] \Rightarrow p \ge 2\sqrt{15} \text{ or } p \le -2\sqrt{15} \\[1em]$

Hence the values of p are $p \le -2\sqrt{15} \text{ or } p\ge 2\sqrt{15}$ .