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# Solve by completing the square: Non-integer solutions

We can use the strategy of completing the square to solve quadratic equations even when the solutions aren't integers. Created by Sal Khan.

## Want to join the conversation?

• There doesn't seem to be any video on the practice: completing the square, and it just totally lost me with everything being added to the squared quadratic. Any tips would help, thanks!
• Try completing the square: intro and the videos associated with it.
• Completing the square method will always work.
• I have been trying at this for hours and The questions only throw more and more hurdles. nothing works for them.
• What exactly do you not understand?
• seriously I agree this is way too confusing to look at. I need some help
• With this trinomial, you need to review the previous lessons on completing the square if you don't understand this. Also, try again tomorrow! You only learn when it's hard and not too confusing!
• i plugged this into the quadratic formula. i got this answer:

-6±√36-36/2
-6±√0/2
-3±√0/2.

Where did I go wrong?
• 4ac = 4(1)(3)=12, so it should be -6±√(36-12)/2. Then the √24=√4*√6=2√6.
• I don't understand why you can't just add 6 to both sides at the beginning. It saves so much time.
• You can, he just splits up stuff into as many individual tasks as possible to make it easy to follow, but you can definitely skip/combine certain steps if you're comfortable with it.
• I am having trouble with this I NEED HELP!
• Why do we have to add +6 to both sides of the 0 = (x+3)^2 - 6 before taking the square root?

Can't we just take the square root without adding the +6 to both sides? Ex. Can't we do something like

√0 = √(x+3)^2 - 6
0 = x+3 +/- 6
+/-6 = x+3
(1 vote)
• In your example you’re not fully taking the square root of both sides. Note that all you did to the -6 was to change it to +/-. If you don’t add 6 to both sides you’d have to do √0 = √((x+3)^2 - 6) to keep both sides of the equation equal. It’s not very easy to take the square root of the whole expression (x+3)^2 - 6. Remember, you can’t distribute the square root. It’s easier just to add the six and then take the square root, getting +/-√ 6 = x+3, which turns to x = -3 +/-√6 once you subtract 3 from both sides.

I hope this helps!
• Why is it +/- sqrt(6) instead of just square root of 6?
(1 vote)
• Example: To find the square root of 9, we need to find a number that, when multiplied by itself, gives us 9. The reason why it is +/-3 is that both positive 3 and -3 give us 9.

-3 * -3 = 9
3 * 3 = 9

This works with all square roots (I think).