Please can someone help me show that the roots of this equation is greater than 1/3
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Please can someone help me show that the roots of this equation is greater than 1/3

[From: ] [author: ] [Date: 12-01-29] [Hit: ]
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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
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