Question

Question made by me.
Find all the solutions for the given equation.
$\sf \implies {x}^{4} + {x}^{2} + 1 = 0$
Thank you :)​

Answer 1

CORRECT QUESTION :-

¤ Find all the solutions for the given equation x4 + x² + 1 = 0.

GIVEN :-

• x4 + x² + 1= 0.

TO FIND :-

• The solutions of the equation .

SOLUTION :-

• Let x² = y.

Therefore new equation becomes,

$\implies \displaystyle \sf \: y ^{2} + y + 1 = 0$

By using the quadratic formula,

$\implies \displaystyle \sf \: y = \dfrac{-b\pm \sqrt {b^{2}-4ac}}{2a}$

• a = 1
• b = 1
• c = 1

Substitute the values in the formula,

$\implies \displaystyle \sf \: y = \dfrac{-1\pm \sqrt {1^{2}-4\times 1}}{2\times 1}$

$\implies \displaystyle \sf \: y = \dfrac{-1\pm \sqrt {1-4}}{2}$

$\implies \displaystyle \sf \: y = \dfrac{-1\pm \sqrt {-3}}{2}$

$\implies \displaystyle \sf \: y = \dfrac{-1+ \sqrt {-3}}{2} , y = \dfrac{-1-\sqrt{-3}}{2}$

Now we assumed that y = x². So,

• when y = (-1 + √-3)/2

⇒x² = (-1 + √-3)/2

⇒x=±√{(-1+ √-3)/2}

Therefore,

⇒x1 = √({-1+ √-3)/2}

⇒x2 = -√{(-1+ √-3)/2}

• when y = (-1 + √-3)/2

⇒x² = (-1- √-3)/2

⇒x= ±√{(-1- √-3)/2}

Therefore,

⇒x3 = √((-1- √-3)/2)

⇒x4=-√((-1- √-3)/2)

Answer 2

$\sf{x^4+x^2+1=0}$

Rewrite the Equation using u = x² and u² = (x²)² = x⁴

$\to\sf{u^2+u+1=0}$

Now solve the Equation u² + u + 1 = 0

• We can solve it using Quadratic Equation

$\to\sf{u_{1,\:2}=\dfrac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}}$

$\to\sf{u_{1,\:2}=\dfrac{-1\pm \sqrt{3}i}{2\cdot \:1}}$

• Separate the Solutions

$\to\sf{\displaystyle u=\frac{-1+\sqrt{3}i}{2\cdot \:1},\:u=\frac{-1-\sqrt{3}i}{2\cdot \:1}}$

The Solutions to the Quadratic Equation are :

$\to\sf{\displaystyle u=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:u=-\frac{1}{2}-i\frac{\sqrt{3}}{2}}$

Substitute back u = x² and solve for x

$\rule{315}{1}$

Solve $\sf{\displaystyle x^2=-\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

• Substitute x  = a + bi

$\to\sf{\displaystyle \left(a+bi\right)^2=-\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

$\to\sf{\displaystyle\left(a^2-b^2\right)+2iab=-\frac{1}{2}+i\frac{\sqrt{3}}{2}}$

• Complex numbers can be equal only if their real and imaginary parts are equal. Rewrite as system of equations :

$\to\begin{bmatrix} \sf{a^2-b^2=-\dfrac{1}{2}}\\\\ \sf{2ab=\dfrac{\sqrt{3}}{2}}\end{bmatrix}$

• On Solving it further, We get

$\to\begin{pmatrix}\sf{a=\dfrac{1}{2}},\:&\sf{b=\dfrac{\sqrt{3}}{2}}\\\\ \sf{a=-\dfrac{1}{2}},\:&\sf{b=-\dfrac{\sqrt{3}}{2}}\end{pmatrix}$

• Substitute back x = a + bi

$\boxed{\to\sf{\displaystyle x=\frac{1}{2}+\frac{\sqrt{3}}{2}i,\:x=-\frac{1}{2}-\frac{\sqrt{3}}{2}i}}$

$\rule{315}{1}$

Solve $\sf{\displaystyle x^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}}$

• Substitute x  = a + bi

$\to\sf{\displaystyle \left(a+bi\right)^2=-\frac{1}{2}-i\frac{\sqrt{3}}{2}}$

$\to\sf{\displaystyle\left(a^2-b^2\right)+2iab=-\frac{1}{2}-i\frac{\sqrt{3}}{2}}$

• Complex numbers can be equal only if their real and imaginary parts are equal. Rewrite as system of equations :

$\to\begin{bmatrix} \sf{a^2-b^2=-\dfrac{1}{2}}\\\\ \sf{2ab=-\dfrac{\sqrt{3}}{2}}\end{bmatrix}$

• On Solving it further, We get

$\to\begin{pmatrix}\sf{a=-\dfrac{1}{2}},\:&\sf{b=\dfrac{\sqrt{3}}{2}}\\\\ \sf{a=\dfrac{1}{2}},\:&\sf{b=-\dfrac{\sqrt{3}}{2}}\end{pmatrix}$

• Substitute back x = a + bi

$\boxed{\to\sf{\displaystyle x=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,\:x=\frac{1}{2}-\frac{\sqrt{3}}{2}i}}$

$\rule{315}{1}$

The Final Solutions Are,

$\boxed{\to\sf{\displaystyle x=\frac{1}{2}+\frac{\sqrt{3}}{2}i,\:x=-\frac{1}{2}-\frac{\sqrt{3}}{2}i,\:x=-\frac{1}{2}+\frac{\sqrt{3}}{2}i,\:x=\frac{1}{2}-\frac{\sqrt{3}}{2}i}}$