Question

x3-12x2+18x-2=0...the roots of the given equation
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Answer 1

Step-by-step explanation:

The given equation is

x³ - 12x² + 18x - 2 = 0

Let us apply the transformation x = y + h in order to remove the second term.

The transformed equation is

(y + h)³ - 12 (y + h)² + 18 (y + h) - 2 = 0

or, y³ + (3h - 12)y² + (3h² - 24h + 18)y + (h³ - 12h² + 18h - 2) = 0

So h = 4 and the equation reduces to

y³ - 30y - 58 = 0 ..... (1)

Let y = u + v

Then y³ = u³ + v³ + 3uv (u + v)

or, y³ = u³ + v³ + 3uvy

or, y³ - 3uvy - (u³ + v³) = 0

Comparing with equation (1), we have

uv = 10

& u³ + v³ = 58

Using the standard form z³ + 3Hz + G = 0 and comparing with (1), we get

H = - 10 and G = - 58

Then uv = - H and u³ + v³ = - G

So u³ = 1/2 {- G + √(G² + 4H³)}

& v³ = 1/2 {- G - √(G² + 4H³)}

Now using values above, we get

u³ = ½ [58 + √{58² - 4 (10³)}]

= ½ {58 + √(3364 - 4000)}

= ½ {58 + √(- 636)}

= ½ (58 + 2√159i)

or, u³ = 29 + √159i

Similarly v³ = 29 - √159i

Since uv = 10, the three values of u and v be

(29 + √159i)^(⅓), (29 + √159i)^(⅓) ω, (29 + √159i)^(⅓) ω²

and (29 - √159i)^(⅓), (29 - √159i)^(⅓) ω², (29 + √159i)^(⅓) ω respectively

where ω is the cubic root of unity.

To find x, we use x = y + h = u + v + h = u + v + 4

This gives the solution for x as

{(29 + √159i)^(⅓) + (29 - √159i)^(⅓) + 4},

{(29 + √159i)^(⅓) ω + (29 - √159i)^(⅓) ω² + 4} and

{(29 + √159i)^(⅓) ω² + (29 - √159i)^(⅓) ω + 4}