If roots of quadratic equation 2x² - kx +k = 0 are real and equal, then find k.
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Answered by
4
Given quadratic equation is= 2x² - kx +k = 0
On comparing with standard form of quadratic equation i.e ax² + bx + c =0,
Here, a = 2 , b= -k, c= k
D(discriminant)= b²-4ac
= (-k)² - 4× 2× k
= k² -8k
= k (k -8)
Since, roots of given equation are real and equal. D= 0
0 = k (k -8)
k = 0 or k-8= 0
k = 0 or k = 8
Hence, required value of k is 0 & 8.
HOPE THIS WILL HELP YOU...
On comparing with standard form of quadratic equation i.e ax² + bx + c =0,
Here, a = 2 , b= -k, c= k
D(discriminant)= b²-4ac
= (-k)² - 4× 2× k
= k² -8k
= k (k -8)
Since, roots of given equation are real and equal. D= 0
0 = k (k -8)
k = 0 or k-8= 0
k = 0 or k = 8
Hence, required value of k is 0 & 8.
HOPE THIS WILL HELP YOU...
Answered by
3
As roots are real and equal then the
Discriminant ⇒ b²-4ac = 0
(-k)²-4(2)(k) = 0
k² - 8k = 0
k² - 8k + 0k + 0 = 0
k(k-8) + 0(k-8) = 0
(k+0)(k-8) = 0
Then k = 8 , 0
:)Hope This Helps!!!
Discriminant ⇒ b²-4ac = 0
(-k)²-4(2)(k) = 0
k² - 8k = 0
k² - 8k + 0k + 0 = 0
k(k-8) + 0(k-8) = 0
(k+0)(k-8) = 0
Then k = 8 , 0
:)Hope This Helps!!!
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