# Is 50 a perfect square?

Apr 12, 2015

50 is not a perfect square.
It does not have an exact square root.

Examples of perfect squares are: Apr 12, 2015

An easy way you could find perfect squares is to memorize the first two, then add 2 to the differences. For example:

1, 4, 9, 16, 25, and 36 are the perfect squares up to ${6}^{2}$.

Now look at the differences.

$4 - 1 = 3$
$9 - 4 = 5$
$16 - 9 = 7$
$25 - 16 = 9$
$36 - 25 = 11$

See a pattern?

So, if you know that ${24}^{2}$ is $576$ and ${25}^{2}$ is $625$, then $\left(625 - 576 + 2\right) + 625 = {26}^{2} = 676$

That is, simply take the difference of two consecutive squares, add $2$, then add it to the higher perfect square.

Jul 19, 2015

Here's an idea rather than an authoritative answer.

It may depend on the context. Normally "No", but possibly "Yes".

#### Explanation:

$50$ is not the perfect square of an integer or rational number. This is what we normally mean by "a perfect square".

It is a square of an irrational, algebraic, real number, namely $5 \sqrt{2}$, therefore you could call it a perfect square in the context of the algebraic numbers.

For example, if you were asked to factor the polynomial $5 {x}^{2} - 1$ you can usefully recognise this as a difference of squares:

$5 {x}^{2} - 1 = {\left(\sqrt{5} x\right)}^{2} - {1}^{2} = \left(\sqrt{5} x - 1\right) \left(\sqrt{5} x + 1\right)$

If recognising $5 {x}^{2}$ as a square means that we consider $5$ as a perfect square being ${\left(\sqrt{5}\right)}^{2}$, then perhaps that's useful.

Another example:

We know that ${x}^{2} + 2 x + 1 = {\left(x + 1\right)}^{2}$ is a perfect square trinomial.

What about $5 {x}^{2} + 10 x + 5$?

It is still the square of a binomial:

$5 {x}^{2} + 10 x + 5 = {\left(\sqrt{5} x + \sqrt{5}\right)}^{2}$

In the context of polynomials, should we reserve the term 'perfect square' for polynomials with rational coefficients?