Find the value of k for the following quadratic equation so that they have two equal roots
Kx(x-2)+6=0
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Given Equation is kx(x - 2) + 6 = 0.
kx^2 - 2kx + 6 = 0.
On comparing with ax^2 + bx + c = 0
a = k, b = -2k, c = 6.
Given that the equation has Equal roots.
D = 0
b^2 - 4ac = 0
(-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
k^2 - 6k = 0
k(k - 6) = 0
k = 0 (or) k = 6.
As we know that k = 0 is not possible.
Therefore the value of k = 6.
Hope this helps!
kx^2 - 2kx + 6 = 0.
On comparing with ax^2 + bx + c = 0
a = k, b = -2k, c = 6.
Given that the equation has Equal roots.
D = 0
b^2 - 4ac = 0
(-2k)^2 - 4(k)(6) = 0
4k^2 - 24k = 0
k^2 - 6k = 0
k(k - 6) = 0
k = 0 (or) k = 6.
As we know that k = 0 is not possible.
Therefore the value of k = 6.
Hope this helps!
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