Math, asked by sunnymehta7347, 29 days ago

find the value of k for which 9x²+2kx+1=0 have equal roots​

Answers

Answered by amansharma264
3

EXPLANATION.

Quadratic polynomial.

⇒ 9x² + 2kx + 1 = 0. have equal roots.

As we know that,

D = Discriminant Or b² - 4ac.

For real and equal roots : D = 0.

⇒ (2k)² - 4(9)(1) = 0.

⇒ 4k² - 36 = 0.

⇒ 4k² = 36.

⇒ k² = 9.

⇒ k = √9.

⇒ k = ± 3.

                                                                                                                       

MORE INFORMATION.

Solution of inequation.

⇒ If a > 0.

(1) = |x| < a Or x² < a² ⇔ - a < x < a.

(2) = |x| > a Or x² > a² ⇔ - a ∪ x > a. Or R - [-a, a].

(3) = If α < β.

(x - α)(x - β) < 0 ⇔ α < x < β.

(x - α)(x - β) ≤ 0 ⇔ α ≤ x ≤ β.

(4) = If α < β.

(x - α)(x - β) > 0 ⇔ x < α ∪ x > β.

(x - α)(x - β) ≥ 0 ⇔ x ≤ α ∪ x ≥ β

Answered by llItzDishantll
10

Answer:

EXPLANATION.

Quadratic polynomial.  

⇒ 9x² + 2kx + 1 = 0. have equal roots.

As we know that,  

D = Discriminant Or b² - 4ac.  

For real and equal roots : D = 0.  

⇒ (2k)² - 4(9)(1) = 0.  

⇒ 4k² - 36 = 0.  

⇒ 4k² = 36.  

⇒ k² = 9.  

⇒ k = √9.  

⇒ k = ± 3.  

                                                                                                                       

EXTRA INFORMATION.

Solution of inequation.

⇒ If a > 0.  

(1) = |x| < a Or x² < a² ⇔ - a < x < a.  

(2) = |x| > a Or x² > a² ⇔ - a ∪ x > a. Or R - [-a, a].  

(3) = If α < β.  

(x - α)(x - β) < 0 ⇔ α < x < β.  

(x - α)(x - β) ≤ 0 ⇔ α ≤ x ≤ β.  

(4) = If α < β.  

(x - α)(x - β) > 0 ⇔ x < α ∪ x > β.  

(x - α)(x - β) ≥ 0 ⇔ x ≤ α ∪ x ≥ β

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