Question

For what value of K ,does the given equation have real and equal roots? (k+1)x2-2(k-1)x-1=0

Answer 1

● We know that, an equation have real and equal roots when its discriminant is equal to 0.

i.e.

b² - 4ac = 0

● From the given equation, we have

a = k+ 1

b = - 2 ( k - 1) = - 2k + 2

c = - 1

● Putting the values

⇒ ( - 2k + 2)² - [ 4 × (k +1) × (-1) ] = 0

⇒ 4k² + 8k + 4 + 4k + 4 = 0

⇒ 4k² + 12k + 8 = 0

⇒ 4k² + ( 8 + 4) k + 8 = 0

⇒ 4k² + 8k + 4k + 8 = 0

⇒ 4k ( k + 2 ) + 4 ( k + 2 ) = 0

⇒ (4k + 4) ( k + 2) = 0

⇒ 4k + 4 = 0

4k = - 4

k = - 1

⇒ k + 2 = 0

k = - 2

So, the equation will have equal and real roots when the value of k will be - 1 or - 2.

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