If the equation x2 + 2(k+2)x + 9k = 0 has equal roots, find k?
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Answered by
17
try hit and trial method. it satisfies for K=1 which gives X=-3 equal roots
Answered by
176
Hi!
x² + 2[ k + 2]x + 9k = 0
a = 1 , b = 2[k+2] , c = 9k
As the equation has equal roots,
b² - 4ac = 0
that is ,
{ 2[k+2] }² - 4 × 9k = 0
4 [ k+2]² - 36k = 0
4 [ k² + 4k + 4 ] - 36k = 0
4 k² + 16k + 16 -36k = 0
4 k² - 20k + 16 = 0
Dividing by 4 ,
k² - 5k + 4 = 0
(k - 4) (k - 1) =0
k = 4 , k = 1
x² + 2[ k + 2]x + 9k = 0
a = 1 , b = 2[k+2] , c = 9k
As the equation has equal roots,
b² - 4ac = 0
that is ,
{ 2[k+2] }² - 4 × 9k = 0
4 [ k+2]² - 36k = 0
4 [ k² + 4k + 4 ] - 36k = 0
4 k² + 16k + 16 -36k = 0
4 k² - 20k + 16 = 0
Dividing by 4 ,
k² - 5k + 4 = 0
(k - 4) (k - 1) =0
k = 4 , k = 1
Anonymous:
correct answer!! Good !
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