If x^2 - 4x + 6 = m(x-4)^2, where m is not equal to 1, has real roots, show that m cannot take any value that is less than 1/3
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x^2 - 4x + 6 = m(x-4)^2 = mx^2 - 8mx + 16m
(m-1)x^2 + (4-8m)x + (16m - 6) = 0
For this quadratic in x to have real roots requires that the discriminant be > 0:
(4 - 8m)^2 - 4(m-1)(16m - 6) > 0
24m - 8 > 0
m > 1/3
(m-1)x^2 + (4-8m)x + (16m - 6) = 0
For this quadratic in x to have real roots requires that the discriminant be > 0:
(4 - 8m)^2 - 4(m-1)(16m - 6) > 0
24m - 8 > 0
m > 1/3