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

Answer 1

$\large\bf\underline {To \: find:-}$

• we need to find the Value of k.

$\large\bf\underline{Given:-}$

kx² - 2√5x + 4 has two real and equal roots

$\huge\bf\underline{Solution:-}$

If the equation has two real and equal roots then , Discriminant = 0

☘ $\bf \star \: {b}^{2} - 4ac = 0$

• ☘ 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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Answer 2

◆ kx²– 2√5 x + 4 = 0

★ Here, a = k, b = – 2√5 x , c = 4

★ Given roots are equal,

◆ D = b² – 4ac = 0

________________________ $\sf$

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