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