Question

for what values of k the equation 9 X square + 6 kX + 4 =0 has equal roots​

Answer 1

Answer:

9x²+6kx+4=0

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

Answer 2

Answer:

k=±2

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