Question

solve (D^4-4D^3+8D^2-8D+4)=0​

Answer 1

Answer:

1 + i & 1 - i

Step-by-step explanation:

Given :

To solve for D if,

$D^4 - 4D^3 + 8D^2 - 8D + 4 = 0$

Solution :

(instead of D, I use x),.

The equation is : $x^4 - 4x^3 + 8x^2 - 8x + 4 = 0$

We

⇒ $x^4 - 4x^3 + 8x^2 - 8x + 4 = 0$

⇒ $x^4 - 4x^3 + 4x^2 + 4x^2 - 8x + 4 = 0$

⇒ $x^2 (x^2 - 4x + 4) + 4(x^2 - 2x + 1) = 0$

⇒ $x^2 (x^2 - 2(2)(x) + (2)^2) + 4(x^2 - 2(1)(x) + (1)^2) = 0$

⇒ $x^2 (x - 2)^2 + 4(x - 1)^2 = 0$

⇒ $[x(x - 2)]^2 + 4(x - 1)^2 = 0$

⇒ $[x^2 - 2x]^2 + 4(x - 1)^2 = 0$

⇒ $[x^2 - 2x + 1 - 1]^2 + 4(x - 1)^2 = 0$

⇒ $[(x - 1)^2 - 1]^2 + 4(x - 1)^2 = 0$

⇒ $[(x - 1)^4 - 2(x - 1)^2 + 1 + 4(x - 1)^2 = 0$

⇒ $[(x - 1)^4 + 2(x - 1)^2 + 1 = 0$

⇒ $[(x - 1)^2 + 1]^2 = 0$

⇒ $(x - 1)^2 + 1 = 0$

⇒ $(x - 1)^2 = -1$

⇒ $x - 1 = ±\sqrt{-1} = i$

⇒ $x = 1 ± i$

∴ The possible values of x (D) are 1 + i & 1 - i  (where i = √-1)

Answer 2

Step-by-step explanation:

i & 1 - i

Step-by-step explanation:

G\yen :

To solve for D if,

D4 - + 8D2 - 8D +4 O

Solution :

(instead of D, I use x),.

The equation is : $4 — 4c3 + 8c2

4=0

We

x4 — 4c3 + 4æ2 4- 4c2 — 8m + 4 c2@2 — 4x +4) + 4(c2 — 2c + 1) = 0

c2(æ2 — 2(2)@) + (2)2) + 4(c2 — + (1)2) = o

[æ2 — 2c]2 + — = 0

— 2æ + I — 112 + — = O

[(æ — — 112 + — = 0

The possible values of x (D) are 1 + i &

1-1 (where i = +1)