Question 767452
Find the value of k so that the equation 3x^2-4kx+k = 0 will have real roots.
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Equation will have 2 equal real roots when the discriminant=0
a=3, b=-4k, c=k
discriminant=b^2-4*a*c=(-4k)^2-4*3*k=16k^2-12k=0
k(16k-12)=0
k=0 (reject)
k=12/16=3/4
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solve for x with k=3/4
3x^2-4kx+k=3x^2-3x+3/4=0
12x^2-12x+3=0
solve for x by quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=12, b=-12, c=3
ans:
x=0.5 (multiplicity 2)