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

Answer:

$k {x}^{2} - 2 \sqrt{5} + 4 = 0$

${b}^{2} - 4ac = 0$

since roots are real and equal...

HERE,

a=k

b= -2√5

c=4

~b^2-4ac=[-2√5]^2 -4×k×4=0

4×5-16k=0

20-16k=0

16k=20

$k = 1.25$

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Answer 2

${ \huge{ \underline{ \underline{ \sf{ \green{GivEn : }}}}}}$

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

${ \huge{ \underline{ \underline{ \sf{ \green{To \: find :}}}}}}$

• Value of k

${ \huge{ \underline{ \underline{ \sf{ \green{SoluTion : }}}}}}$

Given,

kx² - 2√5x + 4

where,

a = k

b = - 2√5

c = 4

We know that,

if the equation has two real and equal roots then , Discriminant will be = 0.

Now, put the value of a, b, c

⟶ b² - 4ac = 0

⟶ (-2√5)² - 4 × k × 4 = 0

⟶ 4 × 5 - 16k = 0

⟶ 20 - 16k = 0

⟶ - 16k = -20

⟶ k = 5/4

Hence,

Value of k is 5/4

__________________________________________________

Again,

Put the value of k in the given equation.

⟶ 5/4x² - 2√5x + 4 = 0

⟶ (5x² - 8√5x + 16)/4 = 0

⟶ 5x² - 8√5x + 16 = 0

⟶ 5x² - 8√5x + 16 = 0

⟶ 5x² - 4√5 - 4√5 + 16 = 0

⟶ √5x(√5x - 4) -4(√5 - 4) = 0

⟶ (√5x - 4)(√5x -4) = 0

⟶ √5x - 4 = 0

⟶ x = 4/√5 Or, ⟶ x = 4/√5

Therefore,

The roots of the given equation are 4/√5 and

4/√5.