Question

Can you solve the equation ${x^4} - 625 = 0$? The answer must be in the form of $a + bi$. Just a few questions:
$1.$How do you find the degree measure to plug into trigonometric or nth roots form?
$2.$ How do you find the total number of solutions?

Answer 1

Hint: In this question, there would be two real solutions and two imaginary solutions. First, we will use the formula ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$. Then we will have to consider two cases according to the equation formed and then have to find the solutions by solving them using the above identity.

Complete step by step answer:
In the above question, it is given that ${x^4} - 625 = 0$.
Now, we have to find the solution of the above equation.
We can also write the above equation as
$ \Rightarrow {\left( {{x^2}} \right)^2} - {\left( {25} \right)^2} = 0$
Now we will use an identity ${a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)$.Therefore,
\[\left( {{x^2} - 25} \right)\left( {{x^2} + 25} \right) = 0\]

Now, there are two cases
Case I: when \[\left( {{x^2} - 25} \right) = 0\]
Now, transpose twenty-five to right hand side
$ \Rightarrow {x^2} = 25$
Taking square root on both sides, we get
$ \Rightarrow x = \pm 5$
Therefore, there are two real values of $x$ as $5\,and\, - 5$.

Case II: when \[\left( {{x^2} + 25} \right) = 0\]
Now, transpose twenty-five to right hand side
$ \Rightarrow {x^2} = - 25$
Taking square root on both sides, we get
$ \Rightarrow x = \pm 5\sqrt { - 1} $
We know that $\sqrt { - 1} = i$.
Therefore, we can also write the above equation as
$ \Rightarrow x = \pm 5i$
Therefore, there are two imaginary values of $x$ as $5i\,and\, - 5i$.
We can also write the above answer as $0 + 5i\,\,and\,\,0\, - 5i$.

Note: In the above solution, the real values are also known as real roots and imaginary values as complex roots. If the power of $x$ was more than that given, then the number of solutions will also be more. It should be remembered that if we are taking the root, then we have to use the constant value because the square can be formed from both the positive as well as negative values.