Question

find the value of k for which the given equation has real and equal roots.

Answer 1

I hope it useful for you

Answer 2

Solution :

Given Quadratic equation ,

5x²-4x+2+k(4x²-2x-1)=0

=> 5x²-4x+2+4kx²-2kx-k=0

=> (5+4k)x²-(4+2k)x+(2-k)=0

Compare this with ax²+bx+c = 0 , we

get ,

a = 5+4k , b = -(4+2k), c = 2-k

Now ,

Discreminant (D) = 0 [ Equal roots ]

=> b² - 4ac = 0

=> [-(4+2k)]² - 4(5+4k)(2-k) = 0

=> 16+4k²+16k -4(10-5k+8k-4k²)=0

=> 16+4k²+16k-40+20k-32k+16k²=0

=> 20k²+4k-24 = 0

=> 5k² + k - 6 = 0

Splitting the middle term,

we get

=>5k² +6k - 5k - 6 =0

=> k( 5k + 6 ) - 1( 5k + 6)=0

=> ( 5k+6)( k - 1 ) = 0

=> 5k + 6 = 0 or k - 1 = 0

=> k = -6/5 or k = 1

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