find the value of k of 9x²+6kx+4=0,when roots are equal
Answers
when roots are equal =》b^2 - 4ac = 0
( 6k )^2 - 4 ( 9) ( 4) = 0
36 k^2 - 144 = 0
36 k^2 = 144
k^2 = 144 ÷ 36
k^2 = 4
k = 2
So the value of k is 2
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SOLUTION :
Option (a) is correct : ± 2/3
Given : 9x² + 6kx + 4 = 0 …………(1)
On comparing the given equation with ax² + bx + c = 0
Here, a = 9 , b = 6k , c = 4
D(discriminant) = b² – 4ac
D = (6k)² - 4 × 9 × 4
D = 36k² - 144
Given equation has equal roots , i e D = 0
36k² - 144 = 0
36(k² - 4) = 0
k² - 4 = 0
k² = 4
k = √4
k = ± 2
On putting k = 2 in eq 1,
9x² + 6(2)x + 4 = 0
9x² + 12x + 4 = 0
Here, a = 9 , b = 12 , c = 4
D = b² - 4ac
D = (12)² - 4 × 9 × 4
D = 144 - 144
D = 0
When D = 0 , then x = - b/2a , x = - b/2a
x = - 12/(2×9)
x = - 12/18 = - ⅔
x = - ⅔
On putting k = - 2 in eq 1,
9x² + 6(-2)x + 4 = 0
9x² - 12x + 4 = 0
Here, a = 9 , b = - 12 , c = 4
D = b² - 4ac
D = (-12)² - 4 × 9 × 4
D = 144 - 144
D = 0
When D = 0 , then x = - b/2a , x = - b/2a
x = -(- 12)/(2×9)
x = 12/18 = ⅔
x = ⅔
Hence, the roots are x = ± ⅔ .
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