Math, asked by shafinshaikh846, 2 months ago

2kx²-3kx+k+1=0 find a k. in quadratic equations​

Answers

Answered by Anonymous
1

Answer:

Step-by-step explanation:

Quadratic equation : x^2 + 3kx + k + 7 = 0

                             ⇒  x^2 + 3kx + ( k + 7 ) = 0

On comparing the given equation with ax^2 + bx + c = 0, we get the following information :

a = 1   ,  b = 3k    ,   c = k + 7

Discriminant is b^2 - 4ac , but for equal & real roots discriminant is 0.

b^2 - 4ac = 0

( 3k )^2 - 4{ 1 × ( k + 7 ) } = 0

9k^2 - 4{ k + 7 } = 0

9k^2 - 4k - 28 = 0

9k^2 - ( 18 - 14 ) k - 28 = 0

9k^2 - 18k + 14k - 28 =  0

9k( k - 2 ) + 14( k - 2 ) = 0

( k - 2 ) ( 9k + 14 ) = 0

                         By Zero Product Rule

k = 2 or k = - 14 / 9

Therefore, value of k is 2 or - 14 / 9

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