What is the square root of 49? Thanks

Jun 20, 2018

$7$

Explanation:

The square root of a number $y$ is a number $x$ with the following property: ${x}^{2} = y$.

So, we are looking for a number that gives $49$ when squared.

Since the number is relatively small, you can go and try to square each number until you obtain $49$:

${1}^{2} = 1$
${2}^{2} = 4$
${3}^{2} = 9$
${4}^{2} = 16$
${5}^{2} = 25$
${6}^{2} = 36$
${7}^{2} = 49$

So, the answer is $7$.

Note: You could observe that $- 7$ satisfies the definition as well, since $- 7$ also gives $49$ when squares. This is up to your definition of square root. In my country, we usually define the square root as the positive number with the property described above.

Without this definition, you may say that $7$ and $- 7$ are both square roots of $49$. I repeat, it is up to how you define the concept. For me, the square root of $49$ is $7$, whereas $7$ and $- 7$ are the two solutions of the equation ${x}^{2} - 49 = 0$.

Jun 20, 2018

$7$

Explanation:

Given: $\sqrt{49}$.

If you studied square numbers, then you should easily notice that $49 = {7}^{2}$, so the expression becomes:

$= \sqrt{{7}^{2}}$

The square and the square root cancel each other out, leaving out just $7$.

But, when you get to higher algebra and learn about quadratic equations, you'll also use the other root, which is $- 7$.

$\sqrt{49} = \pm 7 = \left\mid 7 \right\mid$

Explanation:

Consider that:

${7}^{2} = {\left(- 7\right)}^{2} = 49$

and so

$\sqrt{49} = \sqrt{{7}^{2}} = \sqrt{{\left(- 7\right)}^{2}} = 7 , - 7$

Oftentimes we write this result as:

$\sqrt{49} = \pm 7 = \left\mid 7 \right\mid$