Question

How do you simplify $\sqrt{128}+\sqrt{32}$?

Answer 1

Hint: Expand the equation first. Use the Law of Radicals to simplify the given expression. The multiplication(dissociative) Law suits the most, which is, $\sqrt[n]{{ab}} = \sqrt[n]{a} \times \sqrt[n]{b}$. Try simplifying the constants under the root Firstly, and then apply the Multiplication Law to dissociate the terms. Evaluate further to get a single term.

Complete step-by-step solution:
The given expression is, $\sqrt{128}+\sqrt{32}$
According to one of the Laws of radicals which is the multiplication Law of Radicals,
States that $\sqrt[n]{{ab}} = \sqrt[n]{a} \times \sqrt[n]{b}$ only if $a,b > 0$.
We can write the respective constant in their square-form as below.
$\Rightarrow (\sqrt{{{2}^{6}}\times 2})+(\sqrt{{{2}^{4}}\times 2})$
By using the above reference our expression can be written as,
$\Rightarrow (\sqrt{{{2}^{6}}}\times \sqrt{2})+(\sqrt{{{2}^{4}}}\times \sqrt{2})$
On further simplification, the square and square root gets canceled.
$\Rightarrow ({{2}^{3}}\times \sqrt{2})+({{2}^{2}}\times \sqrt{2})$
$\Rightarrow 8\sqrt{2}+4\sqrt{2}$
Adding the constants in Infront of the radicals Since the radicands are the same.
$\Rightarrow 12\sqrt{2}$
$\therefore \sqrt{128}+\sqrt{32}$ on simplification gives $12\sqrt{2}$.

Note: The Law of Radicals is derived from the Laws of exponents. The expression $\sqrt[n]{a}$, n is called index,$\sqrt{{\text{ }}}$ is called radical, and $a$ is called the radicand. $\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}$The left side of the equation is known as radical form and the right side is exponential form. A Radical represents a fractional exponent in which the numerator and the denominator of the radical contains the power of the base and the index of the radical, respectively.
Memorizing all the Laws of radicals will help to solve any type of same model questions easily. It is important to reduce a radical to its simplest form using the Laws of Radicals for multiplication, division, raising an exponent to an exponent, and taking a radical of a radical makes the simplification process for radicals much easier.