Please help with this equation!! sqrt(x^2-3x)=2 is the answer x=1 or x=4, or both?

Question

anonymous · Answer

$$\sqrt{x^2-3x}=2$$

anonymous · Answer

@binks

jim_thompson5910 · Answer

Let's check each one individually

anonymous · Answer

1 doesn't work right? it's only four, right?

jim_thompson5910 · Answer

Checking x = 1:


$$\large \sqrt{x^2-3x}=2$$


$$\large \sqrt{(1)^2-3(1)}=2$$


$$\large \sqrt{1-3(1)}=2$$


$$\large \sqrt{1-3}=2$$


$$\large \sqrt{-2}=2$$


$$\large \sqrt{-1*2}=2$$


$$\large \sqrt{-1}*\sqrt{2}=2$$


$$\large i\sqrt{2}=2$$


That's FALSE. So x = 1 is NOT a solution.

anonymous · Answer

I was trying to solve the equation instead of plugging in the values! I'm so dumb. You've been very helpful jim thompson.

jim_thompson5910 · Answer

Checking x = 4:


$$\large \sqrt{x^2-3x}=2$$


$$\large \sqrt{(4)^2-3(4)}=2$$


$$\large \sqrt{16-3(4)}=2$$


$$\large \sqrt{16-12}=2$$


$$\large \sqrt{4}=2$$


$$\large 2=2$$


That's TRUE. So x = 4 is defintinitely a solution.

jim_thompson5910 · Answer

Well you should have solved first to get list of potential solutions

jim_thompson5910 · Answer

After you solve, you test each potential solution to see if it is a true solution or not.

anonymous · Answer

You're right! I will do that from now on