Question

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.

Answer 1

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

$x=plusminus \sqrt{a}$

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:
$x^{2} =49 \sqrt{49}=7 or -7$
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, $7^{2} =49$ and $(-7)^{2}=49$. 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 $x^{2} =0$ and $x= \sqrt{0}$. 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 $x^{2} =-49$ which gives us $x= \sqrt{-49}$. 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.