Question

If a-b,a,a+b are zeroes of x^3-6x^2+8x,find the value of b?
Plz help

Answer 1

hey dude !!!!!

The general cubic equation is

a{x}^{3} + b {x}^{2} + cx + d = 0ax

3

+bx

2

+cx+d=0

We know that,

Sum of the roots =

\frac{ - b}{a}

a

−b

Also,

Product of the roots =

\frac{ - d}{a}

a

−d

The given equation is

{x}^{3} - 6 {x}^{2} + 8x = 0x

3

−6x

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}−

1

−6

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}

a

−d

Hence, α³ - αβ² = 0

Or, α³ = αβ²

Or, α² = β²

From 1, β² = 2² = 4

So, β = ±√4 = ±2

hope this helps you

Answer 2

Answer:b=2

Explanation: