Question

Find the value of P for which the following quadratic equation has real and equal roots: x² +2(P-1) x + (P+5)= 0​

Answer 1

Step-by-step explanation:

${b}^{2} - 4ac = 0 \\ {(2p - 2)}^{2} - 4(p + 5) = 0 \\ 4 {p}^{2} - 8p + 4 - 4p - 20 = 0 \\ 4 {p}^{2} -12 p - 16 = 0 \\ {p}^{2} - 3p - 4 = 0 \\ {p}^{2} - 4p + p - 4= 0 \\ p(p - 4) + 1(p - 4) = 0 \\ (p + 1)(p - 4) = 0 \\ p = 4 \: or \: p = - 1$

Answer 2

Answer:

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