Question

the equation x²-kx+16=0 has two equal roots. the equation x²+px+(k-7) =0. both k>0 and p>0. find the value of p​

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$so \: here \: \\ \: quadratic \: \: equation \: \: x {}^{2} - kx + 16 = 0 \\ let \: its \: two \: roots \: be \: \alpha \: and \: \beta \\ \\ accordingly \: here \\ \alpha = \beta \: \: \: \: \: (1) \\ \\ comparing \: this \: equation \: with \\ ax {}^{2} + bx + c = 0 \\ we \: get \\ \\ a = 1 \: , \: b = - k \: , \: c = 16 \\ \\ we \: know \: that \\ \alpha + \beta = \frac{ - b}{a} \\ \\ \alpha + \alpha = \frac{ - ( - k)}{1} \\ \\ = \frac{k}{1} \\ \\ 2 \alpha = k \\ \\ \alpha = \frac{k}{2} \: \: \: \: \: \: \: (2) \\ \\ similarly \\ \\ \alpha \beta = \frac{c}{a} \\ \\ \alpha \times \alpha = \frac{16}{1} \\ \\ \alpha {}^{2} = 16 \\ \\ ( \frac{k}{2} ) {}^{2} = 16 \\ \\ \frac{ k {}^{2} }{4} = 16 \\ \\ k {}^{2} = 64 \\ \\ taking \: square \: roots \: on \: both \: sides \\ we \: have \: then \\ \\ k = ± \: 8$

$but \: as \: here \: \\ k > 0 \\ k = - 8 \: \: is \: absurd \\ \\ thus \: then \\ k = 8$