# Question #bc307

Aug 3, 2015

${x}^{2} + 36$ has no zeros in the set of Real numbers.

#### Explanation:

The is no Real number whose square is $- 36$.
In many textbooks, this kind of equation is said to have "no solution until the author gets around to introducing $\sqrt{- 1}$.

(The square of a positive is positive and the square of a negative is positive.)

We have to invent new kinds of numbers to get a zero for ${x}^{2} + 36$.

Don't think of it as a big deal. For many centuries, the only "numbers" were the counting numbers. There were these things called "ratios of numbers" but for a long time they were not thought of as "numbers".
We need them to solve equations like $3 x - 5 = 0$

Similarly, ratios (fractions) of whole numbers will not allow us to solve ${x}^{2} - 2 =$, so we invent (or discover) a new kind of number (irrational) to solve that equation.

In a similar way, solving ${x}^{2} + 36 = 0$ is not possible if we stick to positive and negative numbers. We need to invent (discover) a new kind os number whose square is $- 36$.

Due to objections to the introduction of this new kind of "number" they were called "Imaginary Numbers"

Historical note

Isaac Newton, one of the founders of physics and calculus, referred to solution to $x + 7 = 0$ as "imaginary". He seemed to believe that sticking a subtraction sign in front of a 'number' did not create a "new kind of number".

Personal note

I still have issues 50 years later from learning (in 2nd grade) that $12 \div 3 = 4$ And $36 \div 2 = 18$ and $12 \div 7$ can't be done.

When I learned about fractions the next year I was angry! I thought (and still think) I should have been taught $12 \div 7$ "can't be done until you learn more math".

Aug 3, 2015

Have a look if it is clear:

#### Explanation:

In the first equation $\left({x}^{2} - 25\right)$ you can substitute two values (the zeros):
$x = 5$
$x = - 5$
and you'll get zero:
${5}^{2} - 25 = 25 - 25 = 0$
and
${\left(- 5\right)}^{2} - 25 = 25 - 25 = 0$

But in the second expression you cannot find two real numbers that makes it equal to zero.
If you try a positive number you'll square it and add to $36$ giving a value different from zero.
If you try a negative number the square will change it into a positive number and, together with $36$, will be different from zero.
It should be better to say that your second expression doesn't have REAL zeros.

Aug 3, 2015

Here's why that's so.

#### Explanation:

${x}^{2} - 25$

can actually be factored as the difference of two squares

$\textcolor{b l u e}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)}$

This means that you can write

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

This expression will be equal to zero when $x - 5 = 0$ and when $x + 5 = 0$, which is why it is said to have two zeros at

$x - 5 = 0 \implies x = \textcolor{g r e e n}{5}$

and

$x + 5 = 0 \implies x = \textcolor{g r e e n}{- 5}$

On the other hand, your second expression doesn't have any zeros that are real numbers.

That happens because the square of any real number, positive or negative, is always positive.

So, if ${x}^{2} \ge 0$ for any $x \in \mathbb{R}$, then the expression ${x}^{2} + 36$ will always be positive as well.

Moreover, this expression will never be equal to zero, because that would imply that ${x}^{2}$ must be negative

${x}^{2} + 36 = 0 \implies \textcolor{red}{\cancel{\textcolor{b l a c k}{{x}^{2} = - 36}}}$

something that cannot happen for real numbers.