x²-2kx+7k-12=0 find the roots of k has real roots
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Step-by-step explanation:
x^2 - 2kx + 7k - 12 = 0
if both roots are real and equals, then the equation has one solution. Then the discrminant must be zero.
==> delta = b^2 - 4ac = 0
==> 4k^2 - 4(1)(7k-12) = 0
==> 4k^2 - 28k + 48 = 0
Divide by 4:
==> k^2 - 7k + 12 = 0
==> (k -4)(k-3) = 0
Then there are two solutions:
k1 = 4
k2= 3
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