Question

If alpha-beta,alpha,alpha+beta are zeros of x3-6x2+8x, then find the value of beta

Answer 1

See the attachment for detail solution
Hope it will help you

Answer 2

The general cubic equation is
$a{x}^{3} + b {x}^{2} + cx + d = 0$
We know that,
Sum of the roots =
$\frac{ - b}{a}$
Also,
Product of the roots =
$\frac{ - d}{a}$
The given equation is
${x}^{3} - 6 {x}^{2} + 8x = 0$
The given roots are (α - β), α and (α + β).

Comparing the given equation with the general cubic equation, a = 1, b = -6, c = 8, d = 0.

According to the question,
Sum of the roots = α - β + α + α + β = 3α =
$- \frac{ - 6}{1}$
Hence, 3α = 6.
So, α = 2. ...(1)

Also, product of the equation = (α - β)α(α + β) = (α² - αβ)(α + β) = α³ + α²β - α²β - αβ² = α³ - αβ² =
$\frac{ - d}{a}$
Hence, α³ - αβ² = 0
Or, α³ = αβ²
Or, α² = β²
From 1, β² = 2² = 4
So, β = ±√4 = ±2.

And that's the answer.