Question

Find the value of k for which the given equation has real and equal roots:

(1) kx² - 6x - 2 = 0 (2) 2x² + kx + 2 = 0.

Answer 1

Heya !!



(1) The given equation is kX² - 6X - 2 = 0


D = [ (-6)² - 4 × K × (-2) ] = (36+8K)




The given equation will have real roots if D _>0




Now,


D _> 0 => 36 + 8K _> => K > -36/8 => K _> -9/2.




(2) The given equation is 2X² + KX + 2 = 0



D = [(K² - 4 × 2 × 2 ) ] = (K² - 16 )




The given equation will have real and equal roots if D_> 0


Now,


D_> 0 => K² - 16 _> 0 => K _> 4 or _< -4.

Answer 2

Hey user !!

So here's u r answer !!

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kx ^2 - 6x -2 = 0

D = b^2 - 4 ac

(- 6 ) ^2 - 4 ( K ) ( - 2 )

36 + 8 k

8K = - 36

k = - 36 / 8

k = - 9 / 2

It has real root
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2 k^2 + kx +2 = 0

D = b^2 - 4 ac

( K ) ^2 - 4 ( 2 ) ( 2 )

k ^2 - 16

k^2 = 16

k = √ 16

k = 4

It has real and equal roots !

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