Question

How do you solve ${{x}^{4}}-3{{x}^{2}}-4=0$?

Answer 1

This type of problem is based on the concept of polynomial equation with degree 4. First, we have to consider the given equation with the variable x. Then, split the middle term of the equation in such a way that it has a common term in both the first and last term. Here, the middle term is $-3{{x}^{2}}$, so we can write $-3{{x}^{2}}$ as the sum of ${{x}^{2}}$ and $-4{{x}^{2}}$. Then, take the common terms from the obtained equation and represent the equation as a product of two functions of ${{x}^{2}}$. Thus, the two functions are the factors of the given equation which should be equated to 0. The, we have to solve both the obtained quadratic equation and find the values of x.

Complete step by step solution:
According to the question, we are asked to find the values of x of ${{x}^{4}}-3{{x}^{2}}-4=0$.
We have been given the equation is ${{x}^{4}}-3{{x}^{2}}-4=0$. ---------(1)
Here, the given equation is with variable x.
Let us now spilt middle term in such a way that the sum of the two terms is equal to -3 and the product of the two terms is equal to -4.
We know that $-4\times 1=-4$ and -4+1=-3.
Let us substitute in equation (1).
We get
${{x}^{4}}+\left( 1-4 \right){{x}^{2}}-4=0$
On using distributive property in the obtained equation, that is, $a\left( b+c \right)=ab+ac$
We get
$\Rightarrow {{x}^{4}}-4{{x}^{2}}+{{x}^{2}}-4=0$
Let us now find the common term.
Here, ${{x}^{2}}$ is the common term from the first two terms and 1 is common from the last two terms.
Therefore, we get
${{x}^{2}}\left( {{x}^{2}}-4 \right)+1\left( {{x}^{2}}-4 \right)=0$.
From the obtained equation, find that $\left( {{x}^{2}}-4 \right)$ is common.
On taking out $\left( {{x}^{2}}-4 \right)$ common from the two terms, we get
$\Rightarrow \left( {{x}^{2}}-4 \right)\left( {{x}^{2}}+1 \right)=0$
Now, we have expressed the given equation as a product of two functions with variable ${{x}^{2}}$.
These two functions are the factors of the equation (1).
Therefore, the factors are $\left( {{x}^{2}}-4 \right)$ and $\left( {{x}^{2}}+1 \right)$.
Since, the equation is equal to 0, the factor will also be equal to 0.
$\Rightarrow \left( {{x}^{2}}-4 \right)=0$ and $\left( {{x}^{2}}+1 \right)=0$.
Now, let us consider $\left( {{x}^{2}}-4 \right)=0$
Let us add 4 on both the sides of the equation.
$\Rightarrow {{x}^{2}}-4+4=0+4$
We know that terms with the same magnitude and opposite signs cancel out. Therefore, we get
${{x}^{2}}=4$
Let us take the square root on both the sides of the expression to get the value of x.
$\Rightarrow \sqrt{{{x}^{2}}}=\sqrt{4}$
We know that $\sqrt{{{x}^{2}}}=\pm x$ and $\sqrt{4}=2$.
$\therefore x=\pm 2$.
Now, let us consider $\left( {{x}^{2}}+1 \right)=0$.
Let us subtract 1 from both the sides of the equation, we get
${{x}^{2}}+1-1=0-1$
We know that terms with the same magnitude and opposite signs cancel out. Therefore, we get
${{x}^{2}}=-1$
Let us take the square root on both the sides of the expression to get the value of x.
$\sqrt{{{x}^{2}}}=\sqrt{-1}$
We know that $\sqrt{{{x}^{2}}}=\pm x$. We get
$x=\pm \sqrt{-1}$.
Also we know that $\sqrt{-1}=i$, where ‘i’ is the complex number.
$\therefore x=\pm i$
Therefore, the values of x in the equation ${{x}^{4}}-3{{x}^{2}}-4=0$ are -2, +2, -i and +i.

Note: We can also solve this question by substituting ${{x}^{2}}=t$. Thus, we get the equation as ${{t}^{2}}-3t-4=0$. Use the quadratic formula, $t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to find the value of t. here, a=1, b=-3 and c=-4. After finding the values of t, substitute t as ${{x}^{2}}$. Take square roots on both the sides and find the values of x.