Question

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

Answer 1

Hint: The given equation is a quadratic equation, to solve this question we should know what polynomial and quadratic equations are. An algorithmic function contacting the numerical values as the coefficient of the unknown variable raised to some power is called a polynomial. The highest exponent in a polynomial equation is known as the degree of the polynomial. A polynomial of degree two is called a quadratic polynomial and a polynomial equation has as many roots as its degree so the given equation will have two roots as it is a quadratic equation.

Complete step by step answer:
We are given –
${x^2} + 3 = 0$
We will take 3 to the right-hand side
${x^2} = - 3$
Square rooting both sides, we get –
$x = \pm \sqrt { - 3} $
We know that $\sqrt { - 1} = i$ where $i$ is called iota.
$ \Rightarrow x = \pm \sqrt 3 i$
Hence the solutions of the equation ${x^2} + 3$ are $\sqrt 3 i$ and $ - \sqrt 3 i$.

Note: The standard form of a quadratic polynomial is $a{x^2} + bx + c = 0$, we find the roots of this form of equations using factorization or completing the square method or the quadratic formula. But in this question, we have a quadratic equation of the form $a{x^2} + c = 0$ that is the value of b is equal to zero that’s why we simply bring c to the other side of the equal to sign and divide both sides by a, then square rooting both the sides of the equation we get the roots. We can also solve the question using the quadratic formula. $\sqrt { - 1} $ is an imaginary number and its value is supposed to be iota $(i)$ so that all the other numbers can be expressed as a multiple of iota, like in this question the solutions are $ \pm \sqrt 3 $ times the imaginary number iota.