Question

How do you simplify the square root of negative $$6$$ and root times square root of negative $$18$$?

Answer 1

Hint: In this question, we have to find out the required value from the given particulars.
We need to first find out the square root of negative $$6$$ . Then we need to find out the square root of negative $$18$$ , then we will multiply the two values. After doing the multiplication we can find out the required solution.

Formula used: We know the formula for imaginary number,
$$i^2=-1$$
i.e., $$i=\sqrt{-1}$$
Where i is the imaginary number.

Complete step-by-step solution:
We need to simplify the square root of negative $$6$$ and root times square root of negative $$18$$ .
First, we need to find out the square root of negative $$6$$ .
If we use, $$i=\sqrt{-1}$$ then we get,
$$\sqrt{-6}=\sqrt{-1\times 6}=\sqrt{-1}.\sqrt{6}=i\sqrt{6}$$.
Now, we need to find out the square root of negative $$18$$ .
$$\sqrt{-18}=\sqrt{-1\times 18}=\sqrt{-1}.\sqrt{18}=i\sqrt{18}$$
Here, the square root of negative $$6$$ end root times square root of negative $$18$$
$$=\sqrt{-6}\times \sqrt{-18}$$
$$=i\sqrt{6}\times i\sqrt{18}$$
$$=i^2\sqrt{6}\sqrt{18}$$
$$=-\sqrt{108}$$ [Since for any real numbers, $$\sqrt{A}.\sqrt{B}=\sqrt{A.B}$$ ]
$$=-\sqrt{2\times 2\times 3\times 3\times 3}$$
$$=-\left(\pm 2\times 3\sqrt{3}\right)$$
$$=\mp 6\sqrt{3}$$

Hence, simplifying the square root of negative $$6$$ and root times square root of negative $$18$$ are either $$6\sqrt{3}$$ or, $$-6\sqrt{3}$$.

Note: For this problem, we need to know what it is i.
A complex number is a number that can be expressed in the form a+bi where a and b are real numbers and i represents the imaginary unit, satisfying the equation $$i^2=-1$$ . Since no real number satisfies this equation, i is called an imaginary number.
Square root:
In mathematics, a square root of a number x is a number y such that, $$y^2=x$$ . In other words, a number y whose square is x.
For example, $$4,-4$$ are square roots of $$16$$ , because $$4^2={\left(-4\right)}^2=16$$ .