Question

Find k so that (k-12)x2+ 2(k-12)x+2=0 has equal roots

Answer 1

Step-by-step explanation:

if the equation have a equal roots then discriminant is zero

(k-12)x2+ 2(k-12)x+2=0

compare with ax² + bx + c = 0

a = (k - 12) , b = 2(k - 12) c=2

let α and β are two roots of the equations

we know that,

α + β = -b/a and ( α x β ) = c / a

here, α = β

α + β = - (2 (k - 12))/ (k - 12)

2 β = -2

β = -1

( α x β ) = c / a

β x β = 2 / (k - 12)

β² (K - 12 ) = 2

put the value of β

(-1)² (k - 12 ) = 2

1 (k - 12 ) = 2

k - 12 = 2

k = 2 + 12

k = 14

2nd method

here we know the value of one of the root of equation, i.e β = -1

put the value of β in equation 1

(k-12)x2+ 2(k-12)x+2=0

(k - 12)(-1)² + 2(k - 12)(-1) + 2 = 0

k - 12 + (-2)(k - 12) + 2 = 0

k - 12 - 2k + 24 + 2 = 0

-k + 12 + 2 = 0

-k + 14 = 0

k = 14