Without solving the following quadratic equations, find the value of 'p' for which the given equations have real and equal roots :

(i) px² - 4x + 3 = 0

(ii) x² + (p - 3)x + p = 0

(i) The given equation is px² - 4x + 3 = 0

Comparing it with ax² + bx + c = 0, we get
a = p , b = -4 , c = 3

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (-4)^2 - 4 \times p \times 3 \\[1em] = 16 - 12p$

For equal roots, discriminant = 0

$\Rightarrow 16 - 12p = 0 \\[1em] \Rightarrow 16 = 12p \\[1em] \Rightarrow p = \dfrac{16}{12} \\[1em] p = \dfrac{4}{3}$

Hence the value of p is $\dfrac{4}{3}$ .

(ii) The given equation is x² + (p - 3)x + p = 0.

Comparing with ax² + bx + c = we obtain,
a = 1 , b = (p - 3) , c = p

$\therefore \text{Discriminant} = b^2 - 4ac \\[1em] = (p - 3)^2 - 4 \times 1 \times p \\[1em] = p^2 + 9 - 6p - 4p \\[1em] = p^2 + 9 - 10p$

For equal roots, discriminant = 0

$\Rightarrow p^2 + 9 - 10p = 0 \\[1em] \Rightarrow p^2 - 10p + 9 = 0 \\[1em] \Rightarrow p^2 - 9p - p + 9 = 0 \\[1em] \Rightarrow p(p - 9) - 1(p - 9) = 0 \\[1em] \Rightarrow (p - 1)(p - 9) = 0 \\[1em] \Rightarrow (p - 1) = 0 \text{ or } p - 9 = 0 \\[1em] \Rightarrow p = 1 \text{ or } p = 9 .$

Hence the value of p is 1, 9.