find the value of k if the equation has two equal roots: kx (x-2) + 6=0
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The given equation can be written in the form of
k {x}^{2} - 2kx + 6 = 0kx2−2kx+6=0
Now it has equal roots that means the Discriminant is 0
D = 0
Also
b2−4ac=04k2−4(k)(6)=04k2−24k=04k(k−6)=0k−6=0k=6
Hence the value of K is 6.
Hope it helps !!!
Please mark as Brainliest.
k {x}^{2} - 2kx + 6 = 0kx2−2kx+6=0
Now it has equal roots that means the Discriminant is 0
D = 0
Also
b2−4ac=04k2−4(k)(6)=04k2−24k=04k(k−6)=0k−6=0k=6
Hence the value of K is 6.
Hope it helps !!!
Please mark as Brainliest.
Answered by
6
kx (x-2) + 6 = 0
(k)x^2 + (-2k)x + 6 = 0
(a)x^2 + (b)x + c = 0. (General form)
For equal roots,
D = 0
b^2 - 4ac = 0. (General form)
(-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
k^2 - 6k = 0
k (k-6) = 0
k = 6, 0
=> 0 is not possible because the resultant equation will not remain quadratic
=> k = 6
Hope this helps... plsss make it the brainliest
(k)x^2 + (-2k)x + 6 = 0
(a)x^2 + (b)x + c = 0. (General form)
For equal roots,
D = 0
b^2 - 4ac = 0. (General form)
(-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
k^2 - 6k = 0
k (k-6) = 0
k = 6, 0
=> 0 is not possible because the resultant equation will not remain quadratic
=> k = 6
Hope this helps... plsss make it the brainliest
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