Square roots are fundamental to algebra and geometry. Whether you’re calculating the side of a square from its area or solving complex quadratic equations, understanding how to manipulate square roots is a vital skill.
In this post, we’ll break down four essential properties of square roots with clear examples to help you master them.
The product of two square roots is equal to the square root of the product of their radicands (the numbers under the root sign). This rule applies as long as both numbers are non-negative.
Property: $\sqrt{a} \times \sqrt{b} = \sqrt{ab}$ (where $a \geq 0, b \geq 0$)
Examples:
Ex 1: $\sqrt{2} \times \sqrt{8} = \sqrt{2 \times 8} = \sqrt{16} = 4$
Ex 2: $\sqrt{45} \times \sqrt{20} = \sqrt{45 \times 20} = \sqrt{900} = 30$
When you multiply a square root by itself (or square a square root), the radical symbol is removed, and you are left with the original number. This is because squaring and finding a square root are inverse operations.
Property: $(\sqrt{a})^2 = \sqrt{a} \times \sqrt{a} = \sqrt{a^2} = a$ (where $a \geq 0$)
Examples:
Ex 1: $\sqrt{2} \times \sqrt{2} = \sqrt{2^2} = 2$
Ex 2: $\sqrt{3} \times \sqrt{3} = \sqrt{3^2} = 3$
Just like multiplication, the division of two square roots can be combined under a single radical. This is incredibly helpful for simplifying fractions containing roots.
Property: $\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}$ (where $a > 0, b > 0$)
Examples:
Ex 1: $\frac{\sqrt{8}}{\sqrt{2}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$
Ex 2: $\frac{\sqrt{243}}{\sqrt{363}} = \sqrt{\frac{243}{363}} = \sqrt{\frac{3 \times 81}{3 \times 121}} = \frac{\sqrt{81}}{\sqrt{121}} = \frac{9}{11}$
If you have a number divided by its own square root, the result is simply the square root of that number. This is essentially a shortcut for “rationalizing the denominator.”
Property: $\frac{a}{\sqrt{a}} = \sqrt{a}$
Examples:
Ex 1: $\frac{2}{\sqrt{2}} = \sqrt{2}$
Ex 2: $\frac{3}{\sqrt{3}} = \sqrt{3}$
A very common mistake is assuming that the sum of two separate square roots is the same as the square root of their sum.
The Reality: $\sqrt{a} + \sqrt{b} \neq \sqrt{a + b}$
Proof by Example:
L.H.S (Left Hand Side): $\sqrt{9} + \sqrt{16} = 3 + 4 = 7$
R.H.S (Right Hand Side): $\sqrt{9 + 16} = \sqrt{25} = 5$
Since $7 \neq 5$, the property does not hold.
The same warning applies to subtraction. You cannot combine two separate radicals into one when subtracting.
The Reality: $\sqrt{a} – \sqrt{b} \neq \sqrt{a – b}$
Proof by Example:
L.H.S: $\sqrt{25} – \sqrt{16} = 5 – 4 = 1$
R.H.S: $\sqrt{25 – 16} = \sqrt{9} = 3$
Here, $1 \neq 3$, so L.H.S $\neq$ R.H.S.
It is tempting to think that the square root “cancels out” the squares inside a sum, but this is a major algebraic error.
The Reality: $\sqrt{a^2 + b^2} \neq a + b$
Proof by Example:
L.H.S: $\sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
R.H.S: $3 + 4 = 7$
Again, $5 \neq 7$, proving this shortcut is incorrect.
The only time you can directly “cancel” the square root with the power of 2 is when the entire sum is being squared as a single unit.
The Rule: $\sqrt{(a + b)^2} = a + b$ (Where $a + b > 0$)
Similar to addition, you cannot simplify the square root of two squared numbers being subtracted by just removing the squares.
The Reality: $\sqrt{a^2 – b^2} \neq a – b$
Proof by Example:
Ex: $\sqrt{5^2 – 4^2} \neq 5 – 4$
If we calculate it: $\sqrt{25 – 16} = \sqrt{9} = 3$, whereas $5 – 4 = 1$.
| Incorrect Assumption | Why it’s wrong |
| $\sqrt{a} + \sqrt{b} = \sqrt{a + b}$ | Sum of roots $\neq$ Root of sum |
| $\sqrt{a} – \sqrt{b} = \sqrt{a – b}$ | Difference of roots $\neq$ Root of difference |
| $\sqrt{a^2 + b^2} = a + b$ | You must solve the sum before taking the root |
In mathematics, the symbol $\sqrt{x}$ refers to the principal square root, which is always non-negative. This leads us to a vital rule:
The result of a square root can never be a negative number. Therefore, $\sqrt{(a-b)^2}$ doesn’t just equal $a-b$; it equals the absolute value of $a-b$.
The equation in the image specifies that $a – b > 0$ (which is the same as saying $a > b$).
Here, the result is simply $a-b$.
Notice the answer is $2$, not $-2$. If we blindly followed the “cancellation” rule without the condition, we would have incorrectly said the answer was $a-b$ (which is $-2$).
The statement $\sqrt{(a – b)^2} = a – b$ is a “conditional identity.” It is only true when $a$ is greater than $b$. If $b$ were greater than $a$, the identity would actually be: