Question

How do you find all the real and complex roots of ${{x}^{4}}-14{{x}^{2}}+45=0$ ?
(a) Factorization
(b) Guessing the roots
(c) Changing the variable
(d) None of the above

Answer 1

Hint: To find all the real and complex roots, we are to try to factorize the equation given as, ${{x}^{4}}-14{{x}^{2}}+45=0$. We are going to analyze the factors of the equation and then try to configure each of them equaling to zero. As it is a 4^th degree equation, we are going to find the 4 roots from this given equation, may it be real and complex.

Complete step-by-step answer:
According to the question, we have been asked to find the methods of finding roots of a cubic equation.
We are trying to factorize the equation, $f(x)={{x}^{4}}-14{{x}^{2}}+45$,
Now, to start with,
${{x}^{4}}-14{{x}^{2}}+45=0$
$\Rightarrow {{x}^{4}}-9{{x}^{2}}-5{{x}^{2}}+45=0$
We know that ${{x}^{2}}-9$ can be taken out as common. Therefore, we get
$\Rightarrow {{x}^{2}}({{x}^{2}}-9)-5({{x}^{2}}-9)=0$
By factorizing, we are getting,
$\Rightarrow ({{x}^{2}}-9)({{x}^{2}}-5)=0$
After more simplification, if we take, $f(x)=0$
Thus, we get,
${{x}^{2}}-9=0$
as ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ , we get
$\Rightarrow {{(x)}^{2}}-{{(3)}^{2}}=0$
$\Rightarrow (x+3)(x-3)=0$
So, we get, $x=3,-3$ .
These are the real roots of the given equation ${{x}^{4}}-14{{x}^{2}}+45=0$.
And again,
${{x}^{2}}-5=0$
$\Rightarrow {{x}^{2}}={{(\sqrt{5})}^{2}}$
$\Rightarrow x=\pm \sqrt{5}$
So, we are getting, $x=\sqrt{5},-\sqrt{5}$
These are also the real roots of the equation ${{x}^{4}}-14{{x}^{2}}+45=0$.
So, all the roots of this equation is said to be, $3,-3,\sqrt{5},-\sqrt{5}$ .
And we can also check that, the hit and trial method can be used to find the solutions of our given problem.
To describe the hit and trial method,
Let us try with the value 3 to start with,
So, we have our equation as, ${{x}^{4}}-14{{x}^{2}}+45$,
putting 3 in the value of x, we get,
$\Rightarrow {{\left( 3 \right)}^{4}}-14.{{\left( 3 \right)}^{2}}+45$
Now, simplifying the value,
\[\Rightarrow 81-14.9+45\]
After more simplification, we will reach the value we want to evaluate.
$\Rightarrow 81-126+45=0$
So, we are getting the value of the function as 0 by putting 3. Then we can conclude that 3 is the root of the equation.
Similarly if we put the values of x as $-3,\sqrt{5},-\sqrt{5}$, we will get the values of the function as zero. Then we can also conclude that $-3,\sqrt{5},-\sqrt{5}$ are the roots of the equation.
Hence, the correct option is, (a) Factorization and (b) Guessing the roots .

Note: In relation to quadratic equations, imaginary numbers (and complex numbers) occur when the value under the radical portion of the quadratic formula is negative. When this occurs, the equation has no roots (zeros) in the set of real numbers. The roots belong to the set of complex numbers, and will be called "complex roots" (or "imaginary roots"). These complex roots will be expressed in the form a + bi. The fact is also known that, ${{i}^{2}}=-1$ . On the other hand, putting the value of ${{x}^{2}}$ as t would give us a quadratic equation and the problem can be solved easily by that.