Math, asked by anjalianjali1234123, 25 days ago

for what value of k , the equation 9x²+6kx+4=0 has equal root​

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Answered by chauhanvikrantsingh
2

Answer:

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Answered by Ranveerx107
4

\underline{\underline{\huge{\gray{\tt{\textbf Answer :-}}}}}

\underline{\underline{\huge{\blue{\tt{\textbf Solution :-}}}}}

  • Option (a) is correct : ± 2/3

: \underline{\underline{\huge{\red{\tt{\textbf 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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