Question

if equation (3x)^2+[27(3^1/k)-15]x+4=0 has equal roots then k is? 1) -2 2)-1/2 3)1/2 4)0

Answer 1

Step-by-step explanation:

here u go

answer should be

k=-1/2

Answer 2

Answer:

k = -1/2

Step-by-step explanation:

The given equation is

$9{x}^{2} + (27. {3}^{ \frac{1}{k} } - 15)x + 4 = 0$

The quadratic equation has equal roots. Therefore the discriminant of the equation is zero.

$d \: = {b}^{2} - 4ac \: = 0$

${(27. {3}^{ \frac{1}{k} }-15) }^{2} - 4.9.4 = 0$

${(27. {3}^{ \frac{1}{k} } - 15)}^{2} = {12}^{2}$

$27. {3}^{ \frac{1}{k}} - 15 = + - 12$

$27. {3}^{ \frac{1}{k} } = 27 \: or \: 3$

${3}^{ \frac{1}{k} } = 1 \: or \: \frac{1}{9}$

${3}^{ \frac{1}{k} } = {3}^{0} or \: {3}^{ - 2}$

$\frac{1}{k} = 0 \: or \: - 2$

$k \: = -\frac{1}{2}$

Since no finite k satisfy 1/k = 0, it is neglected.