Class 11Mathematics Q1 (Quadratic equation)

Q. 1 Solve the given quadratic equation

(x−1)3+8=0,     given that cube roots of unity are 1,ω,ω2.

Solution:

Step 1:

(x−1)3+8=0

(x−1)3=−8

Step 2:

−8=8⋅(−1).

Now, −1 can be written in exponential form as: −1=eiπ.

So,

(x−1)3=8eiπ

Step 3:

The cube roots of 8eiπ are:

x−1=83⋅eiπ/3,83⋅ei(π/3+2π/3),83⋅ei(π/3+4π/3)

Since 83=2, we get:

x−1=2ei(π/3),2ei(π),2ei(5π/3)

Step 4: Simplify:

1. x−1=2eiπ/3=2(cos⁡π3+isin⁡π3)=2(12+i32)=1+i3 ⇒ x=2+i3
2. x−1=2eiπ=2(−1)=−2 ⇒ x=−1
3. x−1=2ei5π/3=2(cos⁡5π3+isin⁡5π3)=2(12−i32)=1−i3 ⇒ x=2−i3

 

Answer: x = −1, x = 2 + i√3, x = 2 − i√3

 

Here formula used:

Euler’s formula: eiθ=cos⁡θ+isin⁡θ

Trigonometric Values

cos⁡π3=12,sin⁡π3=32