## kx² – 2√5 x + 4 = 0. In this eq, find k such that the roots are real and equal and thus find its roots

Question

kx² – 2√5 x + 4 = 0. In this eq, find k such that the roots are real and equal and thus find its roots

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4 weeks 2021-09-22T00:02:30+00:00 2 Answers 0 views 0

1. ◆ kx²– 2√5 x + 4 = 0

★ Here, a = k, b = – 2√5 x , c = 4

★ Given roots are equal,

◆ D = b² – 4ac = 0

________________________ ### Solution:–

⇒ (-2√5)² – 4 × k × 4 = 0

⇒ 4 × 5 – 16k = 0

⇒ 20 – 16k = 0

⇒ – 16k = -20

⇒ k = 20/16

⇒ k = 5/4

### Hence:–

The required value of K = 5/4

2. • we need to find the Value of k. kx² – 2√5x + 4 has two real and equal roots If the equation has two real and equal roots then , Discriminant = 0 • ☘ Equation :- kx² – 2√5x + 4

where,

• a = k
• b = – 2√5
• c = 4
• b² – 4ac = 0

➛ (-2√5)² – 4 × k × 4 = 0

➛ 4 × 5 – 16k = 0

➛ 20 – 16k = 0

➛ – 16k = -20

➛ k = 20/16

➛ k = 5/4

Hence,

• ❥ Value of k is 5/4

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Now,

• Equation :- kx² – 2√5x + 4

⚘ Putting value of k.

➛ 5/4x² – 2√5x + 4

➛ (5x² – 8√5x + 16)/4

➛ 6x² – 8√5x + 16

Now,

Finding roots of the equation :- 6x² – 8√5x + 16 By Middle term splitting method .

➛ 5x² – 8√5x + 16

➛ 5x² – 4√5 – 4√5 + 16

➛ √5x(√5x – 4) -4(√5 – 4)

➛ (√5x – 4)(√5x -4)

➛ √5x – 4 = 0

➛ x = 4/√5 or x = 4/√5

hence ,

❥The roots are 4/√5 and 4/√5

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