Question

Find the value of k so that it has two equal roots 4x2 - 2(k+1)x +(k+4:

Answer 1

Discriminant

Let a quadratic equation be ax² + bx + c = 0

Discriminant Δ = b² - 4 ac

Conditions

For equal roots

b² = 4 ac

For real roots

b² > 4 ac

For imaginary roots

b² < 4 ac

Given they have equal roots .

Comparing 4 x² - 2 ( k + 1 ) x + ( k + 4 ) with a x² + b x + c = 0

a = 4

b = - 2 ( k + 1 )

c = ( k + 4 )

b² = 4 ac

⇒ ( - 2 ( k + 1 ) )² = 4 . 4 ( k + 4 )

⇒ 4 ( k + 1 )² = 16 ( k + 4 )

⇒ ( k + 1 )² = 4 ( k + 4 )

⇒ k² + 1 + 2 k  = 4 k + 16

⇒ k² - 2 k - 15 = 0

⇒ k² - 5 k + 3 k - 15 = 0

⇒ k ( k - 5 ) + 3 ( k - 5 ) = 0

⇒ ( k - 5 )( k + 3 ) = 0

Either k - 5 = 0

Or k + 3 = 0

Either k = 5

Or k = - 3

Answer 2

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