The equation b is the one that can be solved by square roots.
In order to solve an equation by square roots, we want the variable squared and in the other side of the equation, a number so we can apply square root on both sides.
In equation a:
[tex]x^2-9x=0\Rightarrow x^2=9x[/tex]If we apply square root here we won't get specific values for x
In equation b:
[tex]x^2-9=0\Rightarrow x^2=9[/tex]Now we can apply square root. remember that square root have two values:
[tex]\sqrt[]{x^2}=\sqrt[]{9}\Rightarrow x=\pm3[/tex]And that give us the two solution to the equation.
Now to solve equation A, we have several forms. Let's factor x from equation A:
[tex]x^2-9x=0\Rightarrow x(x-9)=0[/tex]now it's clear that one of the solution for this equation is x = 0, because if x = 0 => 0(0-9) = 0
Now The other solution will be when x - 9 = 0 Then
[tex]x-9=0\Rightarrow x=9[/tex]The solution to equation a is x = (0,9)
THe solution to equation b is x= (-3,3)