Question

Without solving the
quadratic equation
Find the value of p for which the
given equation has
real and equal rools.
X square +(p-3) x + p = 0 .

Answer 1

Answer:

since equation has real and equal roots then value of D is 0.

b^2-4ac=0

$(p - 3) { }^{2} - 4 \times 1 \times p = 0$

$p {}^{2} + 9 - 6p - 4p = 0 \ \\ p { }^{2} + 9 - 10p = 0 \\ p { }^{2} - 10p + 9 = 0 \\ p {}^{2} - p - 9p + 9 = 0 \\ p(p - 1) + 9(p - 1) = 0 \\ (p + 9)(p - 1) = 0 \\ p = - 9 \: and \: p = 1$