For the equation x2 = a, describe the values of a that will result in two real solutions, one real solution, and no real solution. complete the explanation below.

The equation given is [tex] x^{2} =a[/tex] where a is some number.
We can solve for x by taking the square root of both sides.

[tex]x=plusminus \sqrt{a}[/tex]

Now let's think through what happens for various values of a.

TWO SOLUTIONS
If a is a positive number the above yields two solutions. Take for example:
[tex] x^{2} =49 \sqrt{49}=7 or -7 [/tex]
There will be two solutions (one positive and one negative) as there are two numbers (here -7 and +7) that when multiplied by themselves give 49. That is, [tex] 7^{2} =49[/tex] and [tex] (-7)^{2}=49 [/tex]. The positive root is called the principal root and the negative root is called the secondary root. This will be the case anytime we take the root of a positive number.

ONE SOLUTION
If a = 0 there is only one solution. That is because [tex]x^{2} =0 [/tex] and [tex]x= \sqrt{0} [/tex]. Zero is neither positive nor negative and it has only one root which is 0 itself. So in this case there is only one solution and it is 0.

NO (REAL) SOLUTIONS
If a is negative we would be taking the square root of a negative number. There is no (real) number that when multiplied by itself gives a negative number. Take for example [tex] x^{2} =-49[/tex] which gives us [tex]x= \sqrt{-49} [/tex]. The square root of -49 is not 7 because (7)(7)=49 which is positive. The square root of -49 is not -7 because (-7)(-7)=49 which is also positive. There is no real number that gives -49 when multiplied by itself. I say "real" numbers because there do exist imaginary/complex numbers but because of the way the questions was asked I imagine you may not know about these yet.