Question

Find value of P in following quaratic equation have real and equal roots 4-(p-2) x + 1=0​

Answer 1

Answer:

your answer is in this attachment

hope it helps :)

Answer 2

Answer:

${b}^{2} - 4ac = 0 \\ {(p - 2)}^{2} - 4 \times 4 \times 1 = 0 \\ {p}^{2} + 4 - 4p - 16 = 0 \\ {p}^{2} - 4p - 12 = 0 \\ {p}^{2} - 6p + 2p - 12 = 0 \\ p(p - 6) + 2(p - 6) = 0 \\ (p + 2)(p - 6) = 0 \\ p = 6 \\ p = - 2$