Determine k so that the equation 2x
2 – kx + 1 = 0 have concident roots
Answers
Answered by
1
Compare given Quadratic equation 2x²-kx+k=0 with ax²+bx+c=0, we get
a = 2,b = -k , c = k,
Discriminant (D)=0
[ Given roots are equal ]
=> b²-4ac=0
=> (-k)²-4×2×k=0
=> k²-8k=0
=> k(k-8)=0
=> k = 0 or k=8
Therefore,
value of k=0 or k=8
Hope this helps u
Answered by
1
ANSWER: k = 2√2
Given 2x^2 - kx + 1
comparing with ax^2 + bx + c = 0
a = 2 b = -k c = 1
discriminant d = 0
then
b^2 - 4ac = 0
(-k)^2 - 4(2)(1) = 0
k^2 - 8 = 0
k^2 = 8
k = √8
k =√4*2
k = √4*√2
k = 2√2
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