Question

Find the value of k , such that the following quadratic equation has equal roots (k−12)2 −2(k−12)+2=0.

Answer 1

Answer:

k =  14

Step-by-step explanation:

equal roots (k−12)x^2 −2(k−12)x+2=0.

For an given equation

ax² + bx + c = 0

roots are equal

if b² = 4ac

in given equation (k−12)x^2 −2(k−12)x+2

a = k-12

b = -2(k-12)

c = 2

(-2(k-12))² = 4×(k-12)×2

=> 4 (k² + 144 - 24k) = 8k - 96

=> k² + 144 - 24k = 2k - 24

=> k² - 26k + 168 = 0

=> k =  {26 ± √((-26)² - (4×168))}/2

=> k =  {26 ± √(676  - 672)}/2

=> k =  {26 ± √4}/2

=> k =  {26 ±  2}/2

=> k =  13 ±  1

or we simply factorise

k² - 26k + 168 = 0

=> k² - 14k -12k + 168 =0

=> k (k-14)-12(k-14) =0

=> (k-14)(k-12) = 0

=> k = 14 or k = 12

k = 14 is only solution as k = 12 not possible as  k-12  = 0 means equation will not be quadratic any more