1. ## why always?

When I do problems involving real roots, such as x = 1, why is it always x = 1 or x = - 1?

For example:

Determine the real roots of each equation:
$\displaystyle (5x^2+20)(3x^2-48)=0$

the real root is x = 4 or x = - 4. Why is it always +/- or how can I identify this issue?

2. Originally Posted by Barthayn
When I do problems involving real roots, such as x = 1, why is it always x = 1 or x = - 1?

For example:
Determine the real roots of each equation:
$\displaystyle (5x^2+20)(3x^2-48)=0$
the real root is x = 4 or x = - 4. Why is it always +/- or how can I identify this issue?
But of course that is not always true for all roots. It is true for square roots.
There are two real square roots of 16, $\displaystyle (4)^2=16~\&~(-4)^2=16$
There are no real square roots of $\displaystyle -16$

There is only one real cube root of 8, $\displaystyle 2^3=8$.
There are no others and we did not use $\displaystyle \pm$.

3. So basically if you can just square root it it will be +/- always? If it is cubic you can only have one root?

4. Originally Posted by Barthayn
So basically if you can just square root it it will be +/- always? If it is cubic you can only have one root?
It is very dangerous to answer a question that contains 'always'.
But yes that is generally true for real numbers.

5. it is because you generated a factor of the form a^2 - b^2 = (a + b)(a - b), the difference of two squares. And why not? They are just exercises, training wheels for a rigorous future math scrimages, if you mind. And multiplicity of roots often comes out handy with some other textbooks . . . .