Question

Solve the given quadratic equation: ${{x}^{4}}+4=0$

Answer 1

Factorize the given polynomial equation into linear factors by using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Equate the linear factors to zero and simplify it to find the roots of the given equation.

Complete step-by-step answer :
We have the equation ${{x}^{4}}+4=0$.
We observe that the equation given to us is a polynomial equation. Polynomial is an expression consisting of variables and coefficients that involves only the operations of addition, subtraction, multiplication or division, and non-negative integer exponents of variables.
Degree of a polynomial is the value of the highest power of degrees of its individual term. We observe that the polynomial given to us is of degree 4.
We have to find the roots of this equation. We will factorize the given equation into linear factors by using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
We can rewrite the equation ${{x}^{4}}+4=0$ as ${{\left( {{x}^{2}} \right)}^{2}}-\left( -4 \right)=0$.
We know that ${{i}^{2}}=-1$, where $i$ is a cube root of unity.
Thus, we can write -4 as $-4={{\left( 2i \right)}^{2}}$.
So, we can rewrite the equation ${{\left( {{x}^{2}} \right)}^{2}}-\left( -4 \right)=0$ as ${{\left( {{x}^{2}} \right)}^{2}}-{{\left( 2i \right)}^{2}}=0$.
We know the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Substituting $a={{x}^{2}},b=2i$ in the above equation, we have ${{\left( {{x}^{2}} \right)}^{2}}-{{\left( 2i \right)}^{2}}=\left( {{x}^{2}}-2i \right)\left( {{x}^{2}}+2i \right)=0$.
Thus, we have factorized the equation ${{x}^{4}}+4=0$ as $\left( {{x}^{2}}-2i \right)\left( {{x}^{2}}+2i \right)=0$.
We will further factorize into linear factors.
Using the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$, we can write ${{x}^{2}}-2i$ as ${{x}^{2}}-2i=\left( x+\sqrt{2i} \right)\left( x-\sqrt{2i} \right).....\left( 2 \right)$.
Similarly, we can write ${{x}^{2}}+2i$ as ${{x}^{2}}+2i={{x}^{2}}-\left( -2i \right)={{x}^{2}}-{{\left( \sqrt{-2i} \right)}^{2}}$.
We can further simplify to write it as ${{x}^{2}}+2i={{x}^{2}}-{{\left( \sqrt{-2i} \right)}^{2}}=\left( x+i\sqrt{2i} \right)\left( x-i\sqrt{2i} \right).....\left( 3 \right)$.
Substituting equation (2) and (3) in equation (1), we have ${{x}^{4}}+4=\left( {{x}^{2}}-2i \right)\left( {{x}^{2}}+2i \right)=\left( x+\sqrt{2i} \right)\left( x-\sqrt{2i} \right)\left( x+i\sqrt{2i} \right)\left( x-i\sqrt{2i} \right)$.
So, we have ${{x}^{4}}+4=\left( x+\sqrt{2i} \right)\left( x-\sqrt{2i} \right)\left( x+{{i}^{\dfrac{3}{2}}}\sqrt{2} \right)\left( x-{{i}^{\dfrac{3}{2}}}\sqrt{2} \right)=0$.
Thus, we have $x=-\sqrt{2i},\sqrt{2i},{{i}^{\dfrac{3}{2}}}\sqrt{2},-{{i}^{\dfrac{3}{2}}}\sqrt{2}$.
Hence, the roots of the equation ${{x}^{4}}+4=0$ are $x=-\sqrt{2i},\sqrt{2i},{{i}^{\dfrac{3}{2}}}\sqrt{2},-{{i}^{\dfrac{3}{2}}}\sqrt{2}$.

Note :To check if the calculated roots are correct or not, we can substitute the value of roots in the equation and check if they satisfy the given equation or not. We can also solve this equation by factorizing the given equation using splitting the middle term or completing the square method.