Find the value of μ for which one root of the quadratic equation μx² - 4x + 8 = 0 is 6 times the other.

Let one root be α. The other is 6α.
Sum of roots = α + 6α = 7α = 4/μ
Product of roots = α(6α) = 6α² = 8/μ
From 7α = 4/μ, α = 4/(7μ)
Substituting in 6α² = 8/μ,
6(4/(7μ))² = 8/μ
6(16/(49μ²)) = 8/μ
96/(49μ²) = 8/μ
96μ = 392μ²
392μ² - 96μ = 0
μ(392μ - 96) = 0
μ = 0 or 392μ = 96
μ = 96/392 = 12/49
Since α = 4/(7μ), if μ = 0, α is undefined.
Therefore μ = 12/49
Let the roots be α and 6α
Sum of roots = 7α = 4/μ
Product of roots = 6α² = 8/μ
From sum of roots, α = 4/(7μ)
Substituting in product of roots:
6(4/(7μ))² = 8/μ
6(16/(49μ²)) = 8/μ
96/(49μ²) = 8/μ
96 = 392μ
μ = 96/392 = 12/49
However, if one root is 6 times the other, then let the roots be α and 6α.
Sum of roots = 7α = 4/μ
Product of roots = 6α² = 8/μ
From the sum of roots, α = 4/(7μ)
Substituting into the product of roots:
6(4/(7μ))² = 8/μ
6(16/49μ²) = 8/μ
96/49μ² = 8/μ
96μ = 392μ²
μ(392μ - 96) = 0
μ = 0 or μ = 96/392 = 12/49
Since μ must be greater than 0 for the quadratic equation to have real roots, μ = 12/49.