Sum and Difference of Square Roots - Rules & Examples

(Video) Sum of Square Roots

by Art of Problem Solving

This video by Art of Problem Solving talks about sum of square roots and shows why you can't split up addition under a radical.

$\large{\text{Summary}}$

With multiplication, we can split up square root products. $$\sqrt{x\times y}=\sqrt x\times\sqrt y$$

But with addition or subtraction, we can't do that. $$\sqrt x\pm\sqrt y\ne\sqrt{x\pm y}$$

Let's look at an example. Suppose we have $\sqrt{25}$ and $\sqrt{144}$. $$\begin{align} \sqrt x+\sqrt y&\ne\sqrt{x+y}\\ \sqrt{25}+\sqrt{144}&\ne\sqrt{25+144}\\ 5+12&\ne\sqrt{169}\\ 17&\ne13 \end{align}$$

So, whether $x$ or $y$ is a positive number or a negative number: $$\boxed{\sqrt{x}\pm\sqrt{y}\text{ is not the same as }\sqrt{x\pm y}}$$

The square root of a number plus the square root of another number is not necessarily equal to the square root of the sum of these numbers. Keep that in mind when looking at the sum of square roots.

One more mistake that can come up is the sum of squares under a square root.

You might think that $\sqrt{8^2+15^2}=8+15$, but this is false. If you simplify each side of the equation, you see they are not equal. $$\begin{align} \sqrt{8^2+15^2}&\ne8+15\\ \sqrt{64+225}&\ne23\\ \sqrt{289}&\ne23\\ 17&\ne23 \end{align}$$

Thus, $$\boxed{\sqrt{x^2\pm y^2}\text{ is not the same as }x\pm y}$$

This topic can feel really tricky, but keep these simple rules in mind when solving problems, and you'll get the hang of it!

You Cannot Split a Sum or Difference Under A Radical Into Two Radicals

If we have a singular product or a quotient under a radical, we are able to split that original radical into separate radicals. Here is an example using cube roots of negative numbers: $$\sqrt[3]{-13x^3} = \sqrt[3]{-13}\sqrt[3]{x^3} = -x\sqrt[3]{13}$$ $$\sqrt[3]{\dfrac{-13}{x^3}} = \dfrac{-\sqrt[3]{13}}{\sqrt[3]{x^3}} = \dfrac{-\sqrt[3]{13}}{x}$$

However, it's a different story when it comes to addition and subtraction under a single radical (which is different than adding and subtracting separate radical terms). If you have a completely simplified sum or a difference under the radical sign, you cannot split it into separate radicals. In general:

For example, if you see this square root $\sqrt{15x^2+x^2}$, you better not even think about rewriting this as: $\sqrt{15x^2}+\sqrt{x^2}.$

Why not? Let's check. Say $x=1.$ $$\begin{align} \sqrt{15x^2+x^2} &\neq \sqrt{15x^2}+\sqrt{x^2}\\ \sqrt{16x^2} &\neq x\sqrt{15}+x\\ 4x&\neq x\sqrt{15}+x\\ 4(1)&\neq (1)\sqrt{15}+(1)\\ 4&\neq\sqrt{15}+1\\ 4&\neq 4.87 \end{align}$$

Here, we can see that these rules can be tricky, so be aware of what's happening under the radical sign! If you try to plug in a value for $x$ in a similar example with subtraction, you'll see similar rules for difference of square roots.

Sums and differences can't be separated into different radicals

Arithmetic under a radical works a lot like arithmetic within parentheses—which is to say, pretty normally, except you have to do it first according to the order of operations. But unlike multiplication and division under radicals, the individual terms in sums and differences can't be separated out into separate radicals.

$$\sqrt{2x + 4} \ne\color{#29A603}{\sqrt{2x} + \sqrt{4}}$$

What this means is that while you can simplify expressions under radicals if you can combine terms, if you can't combine terms, you might have to just treat a whole radical as one unit.

Uncombine-able Radicals

Sometimes, the expression under a radical won't be made of terms that can combine. For example, in the expression: $$x + \sqrt{2x^2 + 7} = 4$$ $2x^2$ and $5$ can't be added together because they are not like terms. In this case, we'd have to find the square of both sides of the equation—which is explained in the linked topic. $$\begin{align} x + \sqrt{2x^2 + 7} = 4 \\ \sqrt{2x^2 + 7} = 4-x \\ \sqrt{2x^2 + 7}^2 =(4-x)^2\\ 2x^2 + 7 =16-8x+x^2\\ x^2 +8x-9=0\\ x=1, -9 \end{align}$$

Combine-able Radicals

Let's say that we want to solve the following expression for $x$: $$\sqrt{5 + 4} + x = 7$$ Since we can add $4$ and $5$ normally on the left hand side, we can just combine them under the radical: $$\sqrt{9} + x = 7$$ And then, since $9$ is a perfect square, we can solve the rest of the equation by using subtraction to move terms around: $$\begin{align} \sqrt{9} + x &= 7 \\\ {\color{#8925AE}{3}} + x &= 7\\\ 3\,{\color{#8925AE}{-\,3}} + x &= 7 \,\color{#8925AE}{-\,3}\\\ x &= \color{#8925AE}{4} \end{align}$$

Let's do another example! Solve for $x$ in the following equation: $$3 = \sqrt{x^2 + 3x^2}$$ Remember, solving for $x$ means you need to get it all alone on one side of the equation. We need to see if we can get rid of the square root.

First, we can combine $x^2$ and $3x^2:$ $$\begin{align} 3 &= \sqrt{x^2 + 3x^2}\\\ 3 &= \sqrt{4x^2} \end{align}$$ Now we need to get rid of the radical. Use what you know from multiplying under radicals and solving equations using division to get $x$ alone on one side of the equation.

$$\begin{align} 3 &= \sqrt{4x^2}\\\ 3 &= \sqrt{4x^2}\\\ 3 &= \sqrt{4} \cdot \sqrt{x^2}\\\ 3 &= 2x\\\ \frac{3}{2} &= x\\ \end{align}$$

The answer is: $\frac{3}{2} = x$.