Find all three solutions for the given equation.
\sf \implies {x}^{3} = - 1⟹x3=−1
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(Class-11th)
Answers
x = -1 , (1 + √3i)/2 , (1 - √3i)/2
Solution :
Given : x³ = -1
To find : Solutions (Roots) of the given equation x³ = -1
We have ,
=> x³ = -1
=> x³ + 1 = 0
=> x³ + 1³ = 0
=> (x + 1)(x² - x•1 + 1²) = 0
=> (x + 1)(x² - x + 1) = 0
Here ,
Two cases arises ;
• x + 1 = 0 OR
• x² - x + 1 = 0
•Case1 :
If x + 1 = 0 , then x = -1
• Case2 :
If x² - x + 1 = 0 , then
This is a quadratic equation in x .
Now ,
Comparing the equation x² - x + 1 = 0 with the general quadratic equation ax² + bx + c = 0 , we have ;
a = 1
b = -1
c = 1
Now ,
The discriminant will be given as ;
=> D = b² - 4ac
=> D = (-1)² - 4•1•1
=> D = 1 - 4
=> D = -3
=> D < 0 , thus the equation x² - x + 1 = 0 will have complex conjugate pair of roots .
Now ,
The roots will be given as ;
x = ( -b ± √D)/2a
x = [-(-1) ± √(-3)]/2•1
x = (1 ± √3i)/2
x = (1 + √3i)/2 , (1 - √3i)/2
Hence ,
All the three roots are ;
x = -1 , (1 + √3i)/2 , (1 - √3i)/2
Answer:-
x = -1 , (1 + √3i)/2 , (1 - √3i)/2
Solution :
Given : x³ = -1
To find : Solutions (Roots) of the given equation x³ = -1
We have ,
=> x³ = -1
=> x³ + 1 = 0
=> x³ + 1³ = 0
=> (x + 1)(x² - x•1 + 1²) = 0
=> (x + 1)(x² - x + 1) = 0
Here ,
Two cases arises ;
• x + 1 = 0 OR
• x² - x + 1 = 0
•Case1 :
If x + 1 = 0 , then x = -1
• Case2 :
If x² - x + 1 = 0 , then
This is a quadratic equation in x .
Now ,
Comparing the equation x² - x + 1 = 0 with the general quadratic equation ax² + bx + c = 0 , we have ;
a = 1
b = -1
c = 1
Now ,
The discriminant will be given as ;
=> D = b² - 4ac
=> D = (-1)² - 4•1•1
=> D = 1 - 4
=> D = -3
=> D < 0 , thus the equation x² - x + 1 = 0 will have complex conjugate pair of roots .
Now ,
The roots will be given as ;
x = ( -b ± √D)/2a
x = [-(-1) ± √(-3)]/2•1
x = (1 ± √3i)/2
x = (1 + √3i)/2 , (1 - √3i)/2
Hence ,
All the three roots are ;
x = -1 , (1 + √3i)/2 , (1 - √3i)/2.....