if 2 x^2-(2 + K)X + K=0 find the value of k for which the roots of the equation are real and equal
Answers
Answered by
1
D = b^2 -4ac
0>(2+k)^2-4 ×k×2
0>4 + k^2+ 4k-8k
0>k^2-4k+4
0>(k-2)^2
0>k-2
k=2
0>(2+k)^2-4 ×k×2
0>4 + k^2+ 4k-8k
0>k^2-4k+4
0>(k-2)^2
0>k-2
k=2
ankitdutta666636:
thank u somuch
Answered by
1
Hi Mate!!!
For real and equal roots Descrimnant ( D) must be equal to zero
D = √ {( 2 + k )² - 4 ( 2 ) ( k )}
D = 0 for real and equal roots.
√{ ( 2 + k )² - 8 k } = 0
squaring both sides
4 + k² + 4k - 8k = 0
k² - 4k + 4 = 0
k = 4 + √ ( 16 - 16 ) / 2 or k = 4 - √ ( 16 - 16 ) / 2
k = 2 or k = 2
Have a nice time...
For real and equal roots Descrimnant ( D) must be equal to zero
D = √ {( 2 + k )² - 4 ( 2 ) ( k )}
D = 0 for real and equal roots.
√{ ( 2 + k )² - 8 k } = 0
squaring both sides
4 + k² + 4k - 8k = 0
k² - 4k + 4 = 0
k = 4 + √ ( 16 - 16 ) / 2 or k = 4 - √ ( 16 - 16 ) / 2
k = 2 or k = 2
Have a nice time...
was it helpful?
I don't know
I use to do like this
sorry, i was asking to anki.
ok
hmm
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