Question

find the value of k for which the roots of the following equation are real and equal Kr(x-2square root 5) +10=0

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Answer 1

SORT WAY:-

kx(x – 2√5 ) + 10 = 0

⇒ kx^2 – 2√5 kx + 10 = 0

Here, a = k, b = – 2√5 k, c = 10

Given roots are equal,

D = b^2 – 4ac = 0

⇒ (–2√5 k)^2 – 4 × k × 10 = 0

⇒ 20k^2 – 40k = 0

⇒ 20k(k – 2) = 0

⇒ k(k – 2) = 0

⇒ k ≠ 0

⇒ k = 2

LONG WAY:-

D= b²-4ac is called discriminant.

The nature of roots depend upon the value of the discriminant D. Since D can be zero, positive or negative.

When D=0, Roots are real and equal.

kx(x-2√5)+10=0

kx²-2√5kx+10=0

Here, a= k, b= -2√5k, c= 10

D= b²-4ac=0

(-2√5k)² - 4×k×10=0

4×5k² -40k =0

20k²-40k=0

20k(k-2)=0

k(k-2)=0

k≠0

k-2=0

k=2

Hence, the value of k is 2.

Answer 2

Answer:

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