Question

find the value of k for which the region 2xsqure -(k-1 )x+8=0

Answer 1

$\underline{\bold{Answer:-}}$


$2x {}^{2} - (k - 1)x + 8 = 0 \\ \\ = > 2x {}^{2} - (kx - x) + 8 = 0 \\ \\ = > 2x {}^{2} - kx + x + 8 = 0 \\ \\ = > 2x {}^{2} + ( - k + 1)x + 8 = 0 \\ \\ using \: \: the \: \: discriminant \\ \\ d = b {}^{2} - 4ac \\ \\ = > d = ( - k + 1) {}^{2} - 4 \times 2 \times 8 \\ \\ = > d = - 63 - 2k + k {}^{2} \\ \\ \\ - 63 - 2k + k {}^{2} = 0 \\ \\ = > k {}^{2} - 2k - 63 = 0 \\ \\ = > k { }^{2} + 7k - 9k - 63 = 0 \\ \\ = > k(k + 7) - 9(k + 7) = 0 \\ \\ = > (k + 7) (k - 9) = 0 \\ \\ = > (k + 7) = 0 \: \: or \: \: (k - 9) = 0 \\ \\ = > k = - 7 \: \: or \: \: k = 9.$








$\textbf{Hope it helps!}$